Claim $a = 0 \text{ m/s}^2$, and the force of friction is $12 \text{ N}$ to the left.

Evidence

• Constant velocity implies $\Delta v = 0$, therefore $a = 0$.
• Since $\Sigma F = ma$, the net force is $0 \text{ N}$.
• $F_{net} = F_{applied} - F_{friction} = 0$.
• $12 \text{ N} - F_{friction}$ $= 0$ $\implies F_{friction}$ $= 12 \text{ N}$.

Reasoning Per Newton's 1st Law, a constant velocity is a state of equilibrium. If the net force is zero, all opposing forces must balance. The applied force to the right must be exactly equaled in magnitude by the force of friction to the left.

Claim Tension $T = M(g + a)$.

Evidence

• Let upward be the positive direction.
• The forces acting on the mass are Tension ($T$) acting upwards and Gravity ($F_g = Mg$) acting downwards.
• $\Sigma F_y = T - F_g = Ma$.
• $T - Mg = Ma$.
• $T = Ma + Mg$.
• Factoring out M: $T = M(a + g)$. Assuming $g = 10 \text{ m/s}^2$, $T = M(a + 10 \text{ m/s}^2)$.

Reasoning Because the object is accelerating upwards, the upward force (Tension) must be greater than the downward force (Gravity). The net force is the difference between the two, which equals mass times acceleration per Newton's Second Law.

Claim Tension $T = M(g - a)$.

Evidence

• Let downward be the positive direction (the direction of acceleration).
• The forces are Tension ($T$) upwards and Gravity ($F_g = Mg$) downwards.
• $\Sigma F_y = F_g - T = Ma$.
• $Mg - T = Ma$.
• $T = Mg - Ma$.
• Factoring out M: $T = M(g - a)$.

Reasoning Because the object is accelerating downwards, the downward force (Gravity) must be greater than the upward force (Tension). We set the direction of acceleration as positive to simplify the math.

Claim $T_{bottom} = m_B (10 \text{ m/s}^2)$ and $T_{top} = (m_A + m_B)(10 \text{ m/s}^2)$.

Evidence

• Since the system is at rest, $a = 0$ for both masses, so $\Sigma F = 0$. Let $g = 10 \text{ m/s}^2$.
• Bottom String (looking at Mass B):
• Forces on B: $T_{bottom}$ (up), $m_B g$ (down).
• $T_{bottom} - m_B g$ $= 0$ $\implies T_{bottom}$ $= m_B (10 \text{ m/s}^2)$.
• Top String (looking at Mass A and B together):
• Treating A and B as a single system of mass $(m_A + m_B)$.
• Forces on system: $T_{top}$ (up), $(m_A + m_B)g$ (down).
• $T_{top} - (m_A + m_B)g$ $= 0$ $\implies T_{top}$ $= (m_A + m_B)(10 \text{ m/s}^2)$.
• Alternatively, analyzing Mass A alone: Forces on A: $T_{top}$ (up), $T_{bottom}$ (down), $m_A g$ (down).
• $T_{top} - T_{bottom} - m_A g$ $= 0$ $\implies T_{top}$ $= m_B g + m_A g$ $\implies T_{top}$ $= (m_A + m_B)g$.

Reasoning In equilibrium, upward forces must balance downward forces. The bottom string only needs to support the weight of the bottom mass. The top string must support the weight of the entire system suspended below it.

Claim $a = \frac{(M - m)g}{M + m}$. Assuming $g = 10 \text{ m/s}^2$, $a = \frac{(M - m) \cdot 10 \text{ m/s}^2}{M + m}$. Rule: One Tension.

Evidence

• Let upward be positive. Acceleration $a > 0$.
• Bottom String (Mass B):
• $\Sigma F_B = m_B a$
• $T_{bottom} - m_B g = m_B a$
• $T_{bottom} = m_B a + m_B g = m_B(a + g)$.
• Top String (System A+B):
• Treat A and B as one mass: $M_{total} = m_A + m_B$.
• $\Sigma F_{sys} = M_{total} a$
• $T_{top} - M_{total} g = M_{total} a$
• $T_{top} = M_{total} a + M_{total} g = (m_A + m_B)(a + g)$.

Reasoning Because the system is accelerating upwards, the tension forces must be greater than the gravitational forces they counteract to provide the necessary net upward force.

Claim Cart C is slowing down. Cart A is speeding up to the right, and Cart B moves to the right at constant velocity.

Evidence

• Cart C: Velocity vector is right ($+v$), acceleration vector is left ($-a$).
• Cart A: Velocity vector is right ($+v$), acceleration vector is right ($+a$).
• Cart B: Velocity vector is right ($+v$), no acceleration vector shown ($a=0$).

Reasoning The cart travels right, but the net force (acceleration) points left. When velocity and acceleration are in opposite directions, the object slows down. When they are in the same direction, the object speeds up. When acceleration is zero, velocity is constant.

Claim $a = g \sin \theta$. Assuming $g = 10 \text{ m/s}^2$, $a = (10 \text{ m/s}^2) \sin \theta$.

Evidence

• At rest, $a = 0$, so net forces in both parallel ($x$) and perpendicular ($y$) directions relative to the ramp are zero.
• y-direction: $F_N - F_g \cos \theta = 0 \implies F_N = mg \cos \theta$.
• x-direction: $F_g \sin \theta - F_{friction}$ $= 0$ $\implies F_{friction}$ $= mg \sin \theta$.
• By definition, maximum static friction is $F_{friction} = \mu_s F_N$.
• Substitute the expressions: $mg \sin \theta = \mu_s (mg \cos \theta)$.
• Divide both sides by $mg \cos \theta$: $\frac{\sin \theta}{\cos \theta} = \mu_s$.
• Therefore, $\mu_s = \tan \theta$.

Reasoning The component of gravity pulling the object down the ramp is balanced perfectly by static friction. The normal force is counteracted by the component of gravity perpendicular to the ramp. The ratio of these two components yields the tangent function, making the coefficient of static friction equal to the tangent of the critical angle.

Claim $T = Mg \sin \theta$.

Evidence

• The box is "resting," meaning it is in equilibrium ($a=0$).
• The forces acting parallel to the ramp are the Tension ($T$) pulling up the incline and the parallel component of gravity ($F_{g\parallel} = Mg \sin \theta$) pulling down the incline.
• $\Sigma F_{\parallel} = 0$.
• $T - Mg \sin \theta = 0$.
• $T = Mg \sin \theta$.

Reasoning For the box to remain stationary on a frictionless ramp, the upward force provided by the string must exactly balance the component of the box's weight pulling it down the slope.

Claim $a = \frac{(M - m)g}{M + m}$. Assuming $g=10$, $a = \frac{(M - m) \cdot 10}{M + m}$. Rule: One Tension.

Evidence

• The system accelerates towards the heavier mass, M. Let the direction of motion for M (downwards) be positive.
• For mass M: $\Sigma F = Ma \implies Mg - T = Ma$
• For mass m (moving upwards): $\Sigma F = ma \implies T - mg = ma$
• Add the two equations together to eliminate T:
• $(Mg - T) + (T - mg) = Ma + ma$
• $Mg - mg = a(M + m)$
• $g(M - m) = a(M + m)$
• $a = \frac{(M - m)g}{M + m}$

Reasoning The net force driving the entire system is the difference in weight between the two masses ($Mg - mg$). This net force accelerates the total mass of the system ($M + m$). We treat the two masses as a single system connected by a single string with uniform tension.

Claim $a = \frac{Mg}{M+m}$. Assuming $g=10$, $a = \frac{M(10)}{M+m}$.

Evidence

• The system is driven by the hanging mass, M.
• For mass m on the table: The only horizontal force is Tension. $T = ma$.
• For hanging mass M: It accelerates downwards. $Mg - T = Ma$.
• Substitute $T$ from the first equation into the second:
• $Mg - (ma) = Ma$
• $Mg = Ma + ma = a(M + m)$
• $a = \frac{Mg}{M + m}$

Reasoning The only external force causing the system to accelerate is the weight of the hanging mass ($Mg$). However, this force must accelerate the combined mass of the entire system ($M + m$). The normal force and weight of m cancel each other out vertically and don't contribute to acceleration.

Claim $T = \frac{Mmg}{M+m}$

Evidence

• Easiest body to use: mass m on the table. The only horizontal force on m is Tension.
• $\Sigma F_m = ma \Rightarrow T = ma$
• Substitute $a = \frac{Mg}{M+m}$ from Ex 1:
• $T = m\left(\frac{Mg}{M+m}\right) = \frac{Mmg}{M+m}$
• Check with hanging mass M: $Mg - T = Ma \Rightarrow T = Mg - Ma = Mg - \frac{M^2 g}{M+m} = \frac{Mmg}{M+m}$ ✓

Reasoning The string only has to accelerate mass m, so $T = ma$. Since $\frac{m}{M+m} < 1$, the tension is always less than $Mg$ — the hanging mass accelerates downward, so the string cannot fully support its weight.

Claim $a = \frac{mg}{M+m}$.

Evidence

• If the roles are reversed, m is hanging and M is on the table.
• The driving force is now the weight of m: $F_{net} = mg$.
• The total mass being accelerated is still $(M + m)$.
• Using Newton's Second Law for the system: $a = \frac{F_{net}}{m_{total}} = \frac{mg}{M+m}$.

Reasoning The logic remains identical to Example 1. The net driving force is the weight of whatever mass is hanging, and it must overcome the inertia of both masses combined.

Claim $a=0$ if $m_{hang}g = M_{ramp}g \sin \theta$, which simplifies to $m_{hang} = M_{ramp} \sin \theta$.

Evidence

• The system is pulled in two competing directions: down the ramp by $M_{ramp}g \sin \theta$, and straight down by $m_{hang}g$.
• For acceleration to be zero, these forces must be equal.
• $m_{hang}g = M_{ramp}g \sin \theta$.
• Dividing both sides by $g$ yields $m_{hang} = M_{ramp} \sin \theta$.
• Part b: The large box will accelerate down the ramp if the force pulling it down the ramp is greater than the force of the hanging mass pulling it up. Therefore, it accelerates down if $M_{ramp}g \sin \theta > m_{hang}g$, or $M_{ramp} \sin \theta > m_{hang}$.

Reasoning The system operates like a tug-of-war. Equilibrium is achieved when the effective pulling force on both sides of the pulley is perfectly balanced. If the parallel component of the ramp mass's weight exceeds the weight of the hanging mass, the system will move in the direction of the ramp.

Claim $a = g \sin \theta$. Assuming $g=10$, $a = (10 \text{ m/s}^2) \sin \theta$.

Evidence

• The only force acting parallel to the ramp is the component of gravity: $F_{g\parallel} = mg \sin \theta$.
• Since there is no friction, this is the net force: $F_{net} = mg \sin \theta$.
• Using Newton's Second Law: $F_{net} = ma$.
• $mg \sin \theta = ma$.
• Divide both sides by $m$: $a = g \sin \theta$.

Reasoning Gravity is the only unbalanced force acting along the plane of motion. The normal force perfectly cancels out the perpendicular component of gravity. Because mass appears on both sides of the $F=ma$ equation, it cancels out, demonstrating that acceleration down a frictionless incline depends only on gravity and the angle, not the mass of the object.

Claim $\mu = 0.30$.

Evidence

• Knowns: $m = 10 \text{ kg}$, $F_{applied} = 40 \text{ N}$, $a = 1.0 \text{ m/s}^2$.
• Calculate Net Force: $F_{net}$ $= ma$ $= (10 \text{ kg})(1.0 \text{ m/s}^2)$ $= 10 \text{ N}$.
• Find Friction: $F_{net} = F_{applied} - F_{friction}$.
  $10 \text{ N}$ $= 40 \text{ N} - F_{friction}$ $\implies F_{friction}$ $= 30 \text{ N}$.
• Find Normal Force: The block is on a flat surface, so $F_N$ $= mg$ $= (10 \text{ kg})(10 \text{ m/s}^2)$ $= 100 \text{ N}$.
• Calculate $\mu$: $F_{friction}$ $= \mu F_N$ $\implies 30 \text{ N}$ $= \mu(100 \text{ N})$ $\implies \mu$ $= 30 \text{ N} / 100 \text{ N}$ $= 0.30$.

Reasoning The applied force must overcome friction to cause acceleration. The difference between the applied force and the actual net force required for the given acceleration gives us the force of friction. We then use the standard friction equation with the calculated normal force.

Claim $F_{push} = 220 \text{ N}$.

Evidence

• Knowns: $m = 20 \text{ kg}$, $\mu = 0.50$, $a = 6.0 \text{ m/s}^2$.
• Find Normal Force: $F_N$ $= mg$ $= (20 \text{ kg})(10 \text{ m/s}^2)$ $= 200 \text{ N}$.
• Find Friction: $F_{friction}$ $= \mu F_N$ $= (0.50)(200 \text{ N})$ $= 100 \text{ N}$.
• Find Net Force: $F_{net}$ $= ma$ $= (20 \text{ kg})(6.0 \text{ m/s}^2)$ $= 120 \text{ N}$.
• Calculate Push: $F_{net}$ $= F_{push} - F_{friction}$ $\implies 120 \text{ N}$ $= F_{push} - 100 \text{ N}$.
• $F_{push} = 120 \text{ N} + 100 \text{ N} = 220 \text{ N}$.

Reasoning To accelerate the box, the push force must provide enough force to overcome kinetic friction AND provide the additional net force required to achieve an acceleration of $6.0 \text{ m/s}^2$. The total push is the sum of these two requirements.

Claim $\mu \approx 0.21$.

Evidence

• Knowns: $m = 40 \text{ kg}$, $\theta = 36.87^\circ$, $T = 200 \text{ N}$, $a = 2.50 \text{ m/s}^2$.
• Resolve Tension:
  $T_x$ $= T \cos(36.87^\circ)$ $= 200 \text{ N}(0.8)$ $= 160 \text{ N}$.
  $T_y$ $= T \sin(36.87^\circ)$ $= 200 \text{ N}(0.6)$ $= 120 \text{ N}$.
• Find Normal Force: The upward component of tension reduces the normal force.
  $F_N$ $= F_g - T_y$ $= (40 \text{ kg})(10 \text{ m/s}^2) - 120 \text{ N}$ $= 400 \text{ N} - 120 \text{ N}$ $= 280 \text{ N}$.
• Find Net Force ($x$-direction): $F_{net}$ $= ma$ $= (40 \text{ kg})(2.50 \text{ m/s}^2)$ $= 100 \text{ N}$.
• Find Friction: $F_{net}$ $= T_x - F_{friction}$ $\implies 100 \text{ N}$ $= 160 \text{ N} - F_{friction}$ $\implies F_{friction}$ $= 60 \text{ N}$.
• Calculate $\mu$: $F_{friction}$ $= \mu F_N$ $\implies 60 \text{ N}$ $= \mu(280 \text{ N})$ $\implies \mu$ $= 60 \text{ N} / 280 \text{ N}$ $\approx 0.214$.

Reasoning Because the rope pulls upwards at an angle, it partially lifts the box, reducing the normal force (and thus friction) compared to a horizontal pull. We must separate the tension into $x$ (driving) and $y$ (lifting) components to solve for the true normal force and net driving force.

Claim $T = 1.26 \times 10^3 \text{ N}$.

Evidence

• Knowns: $m = 1050 \text{ kg}$, $a = 1.20 \text{ m/s}^2$.
• Since there is no friction, the tension is the only horizontal force.
• $F_{net} = ma = T$.
• $T = (1050 \text{ kg})(1.20 \text{ m/s}^2) = 1260 \text{ N}$.

Reasoning With friction ignored, the cable's tension provides 100% of the net force required to accelerate the car's mass according to Newton's Second Law.

Claim Average Force $\approx -3437.5 \text{ N}$ (or $3437.5 \text{ N}$ opposing motion).

Evidence

• Knowns: $m = 1100 \text{ kg}$, $t = 8.0 \text{ s}$, $v_f = 0 \text{ m/s}$.
• Convert velocity to m/s: $v_i$ $= 90 \text{ km/hr} \times \frac{1000 \text{ m}}{3600 \text{ s}}$ $= 25 \text{ m/s}$.
• Find acceleration: $a$ $= \frac{v_f - v_i}{t}$ $= \frac{0 \text{ m/s} - 25 \text{ m/s}}{8.0 \text{ s}}$ $= -3.125 \text{ m/s}^2$.
• Find Force: $F_{net}$ $= ma$ $= (1100 \text{ kg})(-3.125 \text{ m/s}^2)$ $= -3437.5 \text{ N}$.

Reasoning First, kinematics must be used to determine the rate of deceleration required to stop the car in the given timeframe. Once acceleration is known, Newton's Second Law gives the constant net force needed to achieve that deceleration.

Claim $T = 12960 \text{ N}$.

Evidence

• Knowns: $m = 1200 \text{ kg}$, $a = 0.80 \text{ m/s}^2$ (upwards), $g = 10 \text{ m/s}^2$.
• Forces: Tension ($T$) upwards, Gravity ($mg$) downwards.
• $\Sigma F = T - mg = ma$.
• $T = ma + mg = m(a + g)$.
• $T$ $= 1200 \text{ kg}(0.80 \text{ m/s}^2 + 10 \text{ m/s}^2)$ $= 1200 \text{ kg}(10.8 \text{ m/s}^2)$ $= 12960 \text{ N}$.

Reasoning To accelerate an object upwards, the lifting force (tension) must support the object's full weight ($mg$) PLUS provide the extra force ($ma$) needed for acceleration.

Claim $a = 3.7 \text{ m/s}^2$ downward.

Evidence

• Knowns: $m = 10 \text{ kg}$, $T = 63 \text{ N}$ (upwards), $F_g$ $= mg$ $= (10 \text{ kg})(10 \text{ m/s}^2)$ $= 100 \text{ N}$ (downwards).
• Because $F_g > T$, the net force is downwards. Let downward be positive.
• $\Sigma F_y = F_g - T = ma$.
• $100 \text{ N} - 63 \text{ N} = (10 \text{ kg})a$.
• $37 \text{ N}$ $= (10 \text{ kg})a$ $\implies a$ $= 3.7 \text{ m/s}^2$.

Reasoning The downward force of gravity is stronger than the upward tension force. This unbalanced force creates a net force downwards, causing the bucket to accelerate downward.

Claim $a = 21 \text{ m/s}^2$. Tension between the 2.0 kg and 3.0 kg blocks: $T_1 = 42 \text{ N}$. Tension between the 3.0 kg and 5.0 kg blocks: $T_2 = 105 \text{ N}$.

Evidence

• Knowns: $m_1 = 2.0 \text{ kg}$, $m_2 = 3.0 \text{ kg}$, $m_3 = 5.0 \text{ kg}$, $P = 210 \text{ N}$, $\mu = 0$.
• Whole system (all three blocks move together): $a$ $= \frac{P}{m_{total}}$ $= \frac{210 \text{ N}}{10.0 \text{ kg}}$ $= 21 \text{ m/s}^2$.
• Isolate the 2.0 kg block (only $T_1$ pulls it): $T_1$ $= m_1 a$ $= (2.0 \text{ kg})(21 \text{ m/s}^2)$ $= 42 \text{ N}$.
• Isolate the 2.0 kg + 3.0 kg blocks together (only $T_2$ pulls them): $T_2$ $= (m_1 + m_2)a$ $= (5.0 \text{ kg})(21 \text{ m/s}^2)$ $= 105 \text{ N}$.
• Check with the 5.0 kg block: $P - T_2$ $= 210 \text{ N} - 105 \text{ N}$ $= 105 \text{ N}$ $= (5.0 \text{ kg})(21 \text{ m/s}^2)$ ✓

Reasoning With no friction, every block shares the same acceleration. Each string only has to accelerate the mass behind it, so tension grows from the back of the train (42 N) toward the front (105 N), and the pull P accelerates all 10 kg.

Claim $a = 6.0 \text{ m/s}^2$ (4.0 kg down, 1.0 kg up). Tension in the rope: $T = 16 \text{ N}$. Rule: One Tension.

Evidence

• Knowns: $M = 4.0 \text{ kg}$, $m = 1.0 \text{ kg}$, $g = 10 \text{ m/s}^2$, frictionless massless pulley.
• Whole system (the net force is the difference in weights): $a$ $= \frac{(M - m)g}{M + m}$ $= \frac{(4.0 \text{ kg} - 1.0 \text{ kg})(10 \text{ m/s}^2)}{5.0 \text{ kg}}$ $= 6.0 \text{ m/s}^2$.
• Isolate the 1.0 kg mass (accelerating up): $T - mg = ma$, so $T$ $= m(g + a)$ $= (1.0 \text{ kg})(10 \text{ m/s}^2 + 6.0 \text{ m/s}^2)$ $= 16 \text{ N}$.
• Check with the 4.0 kg mass (accelerating down): $Mg - T$ $= 40 \text{ N} - 16 \text{ N}$ $= 24 \text{ N}$ $= (4.0 \text{ kg})(6.0 \text{ m/s}^2)$ ✓

Reasoning One rope over a frictionless pulley has one tension on both sides. The tension must be more than the light mass's weight (10 N) to lift it, but less than the heavy mass's weight (40 N) so it can fall, and 16 N fits between them. The mount holding the pulley pulls up with $2T = 32 \text{ N}$, less than the 50 N total weight, because the system's center of mass is accelerating downward.

Claim $a = 3.0 \text{ m/s}^2$. Tension between the 4.0 kg and 2.0 kg blocks: $T_1 = 12 \text{ N}$. Pull: $P = 30 \text{ N}$.

Evidence

• Knowns: $m_1 = 4.0 \text{ kg}$, $m_2 = 2.0 \text{ kg}$, $m_3 = 3.0 \text{ kg}$, $m_4 = 1.0 \text{ kg}$, $T_2 = 18 \text{ N}$, $T_3 = 27 \text{ N}$, $\mu = 0$.
• Isolate the 3.0 kg block (it sits between the two known tensions): $T_3$ pulls it forward and $T_2$ pulls it back, so $T_3 - T_2 = m_3 a$.
• $a$ $= \frac{T_3 - T_2}{m_3}$ $= \frac{27 \text{ N} - 18 \text{ N}}{3.0 \text{ kg}}$ $= 3.0 \text{ m/s}^2$.
• Isolate the 4.0 kg block (only $T_1$ pulls it): $T_1$ $= m_1 a$ $= (4.0 \text{ kg})(3.0 \text{ m/s}^2)$ $= 12 \text{ N}$.
• Whole system (all four blocks move together): $P$ $= m_{total} a$ $= (10.0 \text{ kg})(3.0 \text{ m/s}^2)$ $= 30 \text{ N}$.
• Check the 2.0 kg block: $T_2 - T_1$ $= 18 \text{ N} - 12 \text{ N}$ $= 6 \text{ N}$ $= (2.0 \text{ kg})(3.0 \text{ m/s}^2)$ ✓   Check the 1.0 kg block: $P - T_3$ $= 30 \text{ N} - 27 \text{ N}$ $= 3 \text{ N}$ $= (1.0 \text{ kg})(3.0 \text{ m/s}^2)$ ✓

Reasoning The block between two known tensions is the key: the difference between those tensions is the net force on that one block, which gives the acceleration. Since every block shares that same acceleration, each remaining tension is just the mass behind that rope times $a$.

Claim $a = 7.0 \text{ m/s}^2$ (1.0 kg slides up the ramp, 4.0 kg falls). Tension $T = 12 \text{ N}$.

Evidence

• Knowns: $m_{ramp} = 1.0 \text{ kg}$, $m_{hang} = 4.0 \text{ kg}$, $\theta = 30^\circ$, $\mu = 0$, $g = 10 \text{ m/s}^2$.
• Hanging weight (drives the system): $W_{hang}$ $= (4.0 \text{ kg})(10 \text{ m/s}^2)$ $= 40 \text{ N}$.
• Weight component down the ramp (resists): $mg\sin\theta$ $= (1.0 \text{ kg})(10 \text{ m/s}^2)\sin 30^\circ$ $= 5.0 \text{ N}$.
• Whole system: $a$ $= \frac{40 \text{ N} - 5.0 \text{ N}}{5.0 \text{ kg}}$ $= 7.0 \text{ m/s}^2$.
• Isolate the 1.0 kg block: $T - mg\sin\theta = ma$ $\implies T$ $= 5.0 \text{ N} + (1.0 \text{ kg})(7.0 \text{ m/s}^2)$ $= 12 \text{ N}$.
• Check with the 4.0 kg mass: $40 \text{ N} - 12 \text{ N}$ $= 28 \text{ N}$ $= (4.0 \text{ kg})(7.0 \text{ m/s}^2)$ ✓

Reasoning Only the component of the block's weight parallel to the ramp opposes the motion; the perpendicular component is balanced by the normal force. Since 40 N is much larger than 5.0 N, the hanging mass wins and drags the block up the incline.

Claim $a = 1.0 \text{ m/s}^2$ toward the 5 kg side. Left tension $T_L = 45 \text{ N}$, right tension $T_R = 33 \text{ N}$.

Evidence

• Knowns: $m_L = 5 \text{ kg}$ (hanging), $m_{table} = 2 \text{ kg}$, $m_R = 3 \text{ kg}$ (hanging), $\mu = 0.5$, $g = 10 \text{ m/s}^2$.
• Weights: $W_L = 50 \text{ N}$, $W_R = 30 \text{ N}$. The 5 kg side is heavier, so the system moves toward it.
• Friction on the 2 kg block: $f$ $= \mu mg$ $= (0.5)(2 \text{ kg})(10 \text{ m/s}^2)$ $= 10 \text{ N}$ (opposes motion, toward the 3 kg side).
• Whole system: $a$ $= \frac{50 \text{ N} - 30 \text{ N} - 10 \text{ N}}{10 \text{ kg}}$ $= 1.0 \text{ m/s}^2$.
• Isolate the 5 kg mass (accelerating down): $W_L - T_L = m_L a$ $\implies T_L$ $= 50 \text{ N} - (5 \text{ kg})(1.0 \text{ m/s}^2)$ $= 45 \text{ N}$.
• Isolate the 3 kg mass (accelerating up): $T_R - W_R = m_R a$ $\implies T_R$ $= 30 \text{ N} + (3 \text{ kg})(1.0 \text{ m/s}^2)$ $= 33 \text{ N}$.
• Check with the 2 kg block: $T_L - T_R - f$ $= 45 - 33 - 10$ $= 2 \text{ N}$ $= (2 \text{ kg})(1.0 \text{ m/s}^2)$ ✓

Reasoning The 20 N difference in hanging weights drives the system, but 10 N of it is used up overcoming friction on the table block, leaving 10 N to accelerate all 10 kg. The two strings have different tensions because the block between them has friction and mass that must be accelerated.

Claim $a = g \sin \theta$. Assuming $g = 10 \text{ m/s}^2$, $a = (10 \text{ m/s}^2) \sin \theta$.

Evidence

• 1. $f = \mu_k F_N$.
• 2. $F_n = mg \cos \theta$.
• 3. $F_{parallel} = mg \sin \theta$.
• 4. Because there is no friction, the only force along the incline is $F_{parallel}$. Therefore, $F_{net} = mg \sin \theta$. By Newton's Second Law, $ma = mg \sin \theta$. The mass cancels, leaving $a = g \sin \theta$.

Reasoning The acceleration down a frictionless plane depends solely on the angle of the incline and gravity, as the mass of the object factors out of the net force equation.

Claim $a = 3.8 \text{ m/s}^2$.

Evidence

• Knowns: $m = 900 \text{ kg}$, $\theta = 25^\circ$, $\mu = 0.0500$. Assume $g = 10 \text{ m/s}^2$.
• Force down the ramp: $F_{g\parallel}$ $= mg \sin \theta$ $= (900 \text{ kg})(10 \text{ m/s}^2)\sin(25^\circ)$ $\approx 9000 \text{ N}(0.4226)$ $\approx 3803.4 \text{ N}$.
• Normal force: $F_N$ $= mg \cos \theta$ $= (900 \text{ kg})(10 \text{ m/s}^2)\cos(25^\circ)$ $\approx 9000 \text{ N}(0.9063)$ $\approx 8156.7 \text{ N}$.
• Force of friction: $F_{friction}$ $= \mu F_N$ $= (0.0500)(8156.7 \text{ N})$ $\approx 407.8 \text{ N}$.
• Net force: $F_{net}$ $= F_{g\parallel} - F_{friction}$ $= 3803.4 \text{ N} - 407.8 \text{ N}$ $= 3395.6 \text{ N}$.
• Acceleration: $a$ $= F_{net} / m$ $= 3395.6 \text{ N} / 900 \text{ kg}$ $\approx 3.77 \text{ m/s}^2$.

Reasoning The net force accelerating the box down the slide is the difference between the component of gravity pulling it down and the kinetic friction resisting its motion.

Claim a) $F_N = 1.7 \times 10^2 \text{ N}$
b) $F_{friction} = 35 \text{ N}$
c) $F_{net} = 65 \text{ N}$
d) $a = 3.3 \text{ m/s}^2$

Evidence

• Knowns: $m = 20 \text{ kg}$, $\theta = 30^\circ$, $\mu = 0.20$. Assume $g = 10 \text{ m/s}^2$.
• a) $F_N$ $= mg \cos \theta$ $= (20 \text{ kg})(10 \text{ m/s}^2)\cos(30^\circ)$ $= 200 \text{ N}(0.866)$ $\approx 173.2 \text{ N}$.
• b) $F_{friction}$ $= \mu F_N$ $= (0.20)(173.2 \text{ N})$ $\approx 34.64 \text{ N}$.
• c) Force down ramp: $F_{g\parallel}$ $= mg \sin \theta$ $= (20 \text{ kg})(10 \text{ m/s}^2)\sin(30^\circ)$ $= 200 \text{ N}(0.5)$ $= 100 \text{ N}$.
  $F_{net}$ $= F_{g\parallel} - F_{friction}$ $= 100 \text{ N} - 34.64 \text{ N}$ $\approx 65.36 \text{ N}$.
• d) $a$ $= F_{net} / m$ $= 65.36 \text{ N} / 20 \text{ kg}$ $\approx 3.268 \text{ m/s}^2$.

Reasoning This problem sequentially breaks down the standard ramp problem process. First find the perpendicular force to determine friction, then find the parallel force to determine the net driving force, and finally use the net force to calculate acceleration.