Question

Three blocks of masses 2 kg, 3 kg, 4 kg are placed as shown in the figure. Coefficient of friction between the surfaces are 0.2, 0.5, 0.0. A horizontal force is applied on 3 kg block

The work done by the friction when the force of 50 newton acts on the 3 kg blocks for 2 second

• A. 54 joule
• B. 108 joule
• C. 216 joule
• D. 512 joule

Answer 1

Answer: (B)

108 joule

Answer 2

The problem involves three blocks stacked on top of each other with different coefficients of friction between the surfaces. A horizontal force of 50 N is applied to the 3 kg block. We need to find the work done by friction in 2 seconds.

Let $m_1 = 2$ kg, $m_2 = 3$ kg, $m_3 = 4$ kg. Coefficients of friction: $\mu_1 = 0.2$ (between 2 kg and 3 kg), $\mu_2 = 0.5$ (between 3 kg and 4 kg), $\mu_3 = 0.0$ (between 4 kg and ground). Applied force $F = 50$ N on the 3 kg block. Time $t = 2$ s. Use $g = 10$ m/s$^2$.

First, determine if the blocks move together or slide relative to each other. Maximum static friction between 2 kg and 3 kg: $f_{s1, max} = \mu_1 m_1 g = 0.2 \times 2 \times 10 = 4$ N. Maximum static friction between 3 kg and 4 kg: $f_{s2, max} = \mu_2 (m_1+m_2) g = 0.5 \times (2+3) \times 10 = 0.5 \times 50 = 25$ N. Maximum static friction between 4 kg and ground: $f_{s3, max} = \mu_3 (m_1+m_2+m_3) g = 0.0 \times (2+3+4) \times 10 = 0$ N.

Assume all blocks move together with acceleration $a$. Total mass $M = 2+3+4 = 9$ kg. $F = M a \implies 50 = 9 a \implies a = \frac{50}{9}$ m/s$^2$. Required friction on 2 kg from 3 kg is $f_1 = m_1 a = 2 \times \frac{50}{9} = \frac{100}{9} \approx 11.11$ N. Since $f_1 > f_{s1, max}$ (11.11 N > 4 N), the 2 kg block slides relative to the 3 kg block. The friction between 2 kg and 3 kg is kinetic friction: $f_{k1} = \mu_1 m_1 g = 4$ N. This force is on 2 kg in the direction of motion and on 3 kg opposite to the relative motion.

Now consider the 2 kg block. The only horizontal force is $f_{k1}$ from the 3 kg block. $f_{k1} = m_1 a_1 \implies 4 = 2 a_1 \implies a_1 = 2$ m/s$^2$.

Now consider the 3 kg and 4 kg blocks. Let's assume they move together with acceleration $a_{23}$. The forces on the combined (3+4) kg system are the applied force $F$, the kinetic friction from the 2 kg block on the 3 kg block ($f_{k1}$ backward), and the friction from the ground on the 4 kg block ($f_3$). Since $\mu_3 = 0$, $f_3 = 0$. The force on the 3 kg block from the 2 kg block is $f_{k1}$ backward. The force on the 4 kg block from the 3 kg block is $f_2$ forward. Consider the combined (3+4) kg system. The external horizontal forces are $F$ and $f_{k1}$ (backward). $F - f_{k1} = (m_2+m_3) a_{23}$ (This is incorrect because $f_{k1}$ is an internal force in this combined system). Let's consider the forces on 3 kg and 4 kg separately. On 3 kg: $F - f_{k1} - f_2 = m_2 a_2$. $50 - 4 - f_2 = 3 a_2 \implies 46 - f_2 = 3 a_2$. On 4 kg: $f_2 - f_3 = m_3 a_3$. Since $f_3 = 0$, $f_2 = 4 a_3$. If 3 kg and 4 kg move together, $a_2 = a_3 = a_{23}$. $f_2 = 4 a_{23}$. $46 - 4 a_{23} = 3 a_{23} \implies 46 = 7 a_{23} \implies a_{23} = \frac{46}{7}$ m/s$^2$. The required friction between 3 kg and 4 kg is $f_2 = 4 a_{23} = 4 \times \frac{46}{7} = \frac{184}{7} \approx 26.29$ N. Since $f_2 > f_{s2, max}$ (26.29 N > 25 N), the 3 kg block slides relative to the 4 kg block. The friction between 3 kg and 4 kg is kinetic friction: $f_{k2} = \mu_2 (m_2+m_1) g$ is incorrect. The normal force between 3kg and 4kg is from the weight of 2kg and 3kg blocks, so $N_2 = (m_1+m_2)g$. $f_{k2} = \mu_2 N_2 = 0.5 \times (2+3) \times 10 = 25$ N. This force is on 3 kg opposite to its motion relative to 4 kg, and on 4 kg in the direction of motion of 3 kg relative to 4 kg.

Now, we have kinetic friction at both interfaces (2-3 and 3-4). Forces on 2 kg: $f_{k1} = 4$ N (forward). $a_1 = \frac{f_{k1}}{m_1} = \frac{4}{2} = 2$ m/s$^2$. Forces on 3 kg: $F = 50$ N (forward), $f_{k1} = 4$ N (backward from 2 kg), $f_{k2} = 25$ N (backward from 4 kg). $F - f_{k1} - f_{k2} = m_2 a_2 \implies 50 - 4 - 25 = 3 a_2 \implies 21 = 3 a_2 \implies a_2 = 7$ m/s$^2$. Forces on 4 kg: $f_{k2} = 25$ N (forward from 3 kg), $f_{k3} = 0$ N (from ground). $f_{k2} - f_{k3} = m_3 a_3 \implies 25 - 0 = 4 a_3 \implies a_3 = \frac{25}{4} = 6.25$ m/s$^2$.

Accelerations: $a_1 = 2$ m/s$^2$, $a_2 = 7$ m/s$^2$, $a_3 = 6.25$ m/s$^2$. Displacements after $t = 2$ s (starting from rest): $d_1 = \frac{1}{2} a_1 t^2 = \frac{1}{2} \times 2 \times 2^2 = 4$ m. $d_2 = \frac{1}{2} a_2 t^2 = \frac{1}{2} \times 7 \times 2^2 = 14$ m. $d_3 = \frac{1}{2} a_3 t^2 = \frac{1}{2} \times 6.25 \times 2^2 = 12.5$ m.

Work done by friction at the 2-3 interface: The friction force on 2 kg is $f_{k1} = 4$ N in the direction of motion. Work done on 2 kg is $W_{f1, 2} = f_{k1} d_1 = 4 \times 4 = 16$ J. The friction force on 3 kg is $f_{k1} = 4$ N opposite to the direction of motion. Work done on 3 kg is $W_{f1, 3} = -f_{k1} d_2 = -4 \times 14 = -56$ J. Total work done by friction at the 2-3 interface is $W_{f1} = W_{f1, 2} + W_{f1, 3} = 16 - 56 = -40$ J.

Work done by friction at the 3-4 interface: The friction force on 3 kg is $f_{k2} = 25$ N opposite to the direction of motion. Work done on 3 kg is $W_{f2, 3} = -f_{k2} d_2 = -25 \times 14 = -350$ J. The friction force on 4 kg is $f_{k2} = 25$ N in the direction of motion. Work done on 4 kg is $W_{f2, 4} = f_{k2} d_3 = 25 \times 12.5 = 312.5$ J. Total work done by friction at the 3-4 interface is $W_{f2} = W_{f2, 3} + W_{f2, 4} = -350 + 312.5 = -37.5$ J.

Work done by friction at the 4-ground interface is 0 since the friction force is 0.

The total work done by friction is the sum of work done at each interface: $W_{friction} = W_{f1} + W_{f2} + W_{f3} = -40 + (-37.5) + 0 = -77.5$ J.

However, the options are positive. The question asks for the work done by the friction. This might mean the magnitude of the work done by the friction forces on the system or the sum of the absolute values of the work done by friction on each block. Let's check the options again. The options are 54, 108, 216, 512. None of these match -77.5.