Question

A small block B of mass m is placed on a plank P of mass M and length $l$, which is placed on a frictionless horizontal floor. Coefficient of friction between the block and the plank is $\mu$. Initially the block is near the right edge of the plank and the setup is at rest. Such a constant horizontal force F is applied on the plank that the block begins to slide on the plank. Find work done by the force F, work done by the frictional force on the block, and work done by the fictional force on the plank, when the block reaches the left edge of the plank.

Answer 1

Work done by F: $\frac{F(F - \mu mg)l}{F - \mu g(M+m)}$, Work done by friction on block: $\frac{\mu^2 m M g^2 l}{F - \mu g(M+m)}$, Work done by friction on plank: $-\frac{\mu mg (F - \mu mg)l}{F - \mu g(M+m)}$

Answer 2

The problem involves a block (B) of mass $m$ on a plank (P) of mass $M$ and length $l$. A force $F$ is applied to the plank, causing the block to slide. We need to find the work done by the applied force $F$, the work done by friction on the block, and the work done by friction on the plank. The floor is frictionless.

1. Forces and Accelerations:

Since the block slides on the plank, the friction acting between them is kinetic friction. The normal force between the block and the plank is $N = mg$. The kinetic friction force is $f_k = \mu N = \mu mg$.

• For the block (B): The only horizontal force acting on the block is the friction force from the plank, $f_k$. This force acts in the direction of the block's acceleration. Assuming the force $F$ is applied to the right, the plank accelerates right. The block tends to lag, so friction on the block is to the right. Using Newton's second law: $f_k = m a_B$ $\mu mg = m a_B$ $a_B = \mu g$ (acceleration of the block)

• For the plank (P): The horizontal forces acting on the plank are the applied force $F$ (to the right) and the friction force from the block, $f_k$ (to the left, opposing the plank's motion relative to the block). Using Newton's second law: $F - f_k = M a_P$ $F - \mu mg = M a_P$ $a_P = \frac{F - \mu mg}{M}$ (acceleration of the plank)

For the block to slide on the plank, $a_P > a_B$. $\frac{F - \mu mg}{M} > \mu g \implies F - \mu mg > M\mu g \implies F > \mu g(M+m)$. This condition is implied by the problem statement that the block "begins to slide".

2. Displacements:

The block starts near the right edge and reaches the left edge of the plank. This means the relative displacement of the block with respect to the plank is $l$ (the block moves $l$ distance backward relative to the plank, or the plank moves $l$ distance forward relative to the block). Let $x_B$ be the absolute displacement of the block and $x_P$ be the absolute displacement of the plank. Both start from rest. Since the plank moves ahead of the block, the relative displacement is $x_P - x_B = l$. Using kinematic equations ($x = \frac{1}{2} a t^2$): $\frac{1}{2} a_P t^2 - \frac{1}{2} a_B t^2 = l$ $\frac{1}{2} (a_P - a_B) t^2 = l$ $t^2 = \frac{2l}{a_P - a_B}$ Substitute the expressions for $a_P$ and $a_B$: $t^2 = \frac{2l}{\left(\frac{F - \mu mg}{M}\right) - \mu g} = \frac{2l}{\frac{F - \mu mg - M\mu g}{M}} = \frac{2Ml}{F - \mu g(M+m)}$

Now, find the absolute displacements: $x_B = \frac{1}{2} a_B t^2 = \frac{1}{2} (\mu g) \left( \frac{2Ml}{F - \mu g(M+m)} \right) = \frac{\mu g Ml}{F - \mu g(M+m)}$ $x_P = \frac{1}{2} a_P t^2 = \frac{1}{2} \left(\frac{F - \mu mg}{M}\right) \left( \frac{2Ml}{F - \mu g(M+m)} \right) = \frac{(F - \mu mg)l}{F - \mu g(M+m)}$

3. Work Done Calculations:

• Work done by the force F ($W_F$): The force $F$ acts on the plank in the direction of its displacement $x_P$. $W_F = F \cdot x_P$ $W_F = F \left( \frac{(F - \mu mg)l}{F - \mu g(M+m)} \right)$

• Work done by the frictional force on the block ($W_{f,B}$): The friction force $f_k = \mu mg$ acts on the block in the direction of its displacement $x_B$. $W_{f,B} = f_k \cdot x_B$ $W_{f,B} = (\mu mg) \left( \frac{\mu g Ml}{F - \mu g(M+m)} \right)$ $W_{f,B} = \frac{\mu^2 m M g^2 l}{F - \mu g(M+m)}$

• Work done by the frictional force on the plank ($W_{f,P}$): The friction force $f_k = \mu mg$ acts on the plank opposite to the direction of its displacement $x_P$. $W_{f,P} = -f_k \cdot x_P$ $W_{f,P} = -(\mu mg) \left( \frac{(F - \mu mg)l}{F - \mu g(M+m)} \right)$

The work done by friction on the block is positive because friction accelerates the block. The work done by friction on the plank is negative because friction opposes the plank's motion.