Question: A block is projected on a rough horizontal surface with velocity $v_0$ at $t=0$. At time $t=t_0$ blo...

Question

A block is projected on a rough horizontal surface with velocity $v_0$ at $t=0$ . At time $t=t_0$ block stops due to friction. Now consider a reference frame 'S' moving with constant velocity $v_0$ in the direction of block. Work done by the friction on the surface from $t=0$ to $t=t_0$ , as observed by reference frame 'S' is:-

Answer 1

Solution

1. Analyze the motion in the Ground Frame (G):

Let the initial velocity of the block be $v_0$ in the positive x-direction.
The block experiences a constant frictional force $f = \mu mg$ (where $\mu$ is the coefficient of kinetic friction and $m$ is the mass of the block) in the negative x-direction.
The acceleration of the block is $a = -f/m = -\mu g$ .
The block stops at time $t=t_0$ . Using the equation of motion $v = u + at$ :
$0 = v_0 + (-\mu g)t_0$
$v_0 = \mu g t_0$ (Equation 1)

2. Identify the Force on the Surface:

The question asks for the work done by friction on the surface.
The block exerts a friction force on the surface. Since the block moves to the right (positive x-direction) relative to the surface, the force of friction exerted by the block on the surface is in the positive x-direction.
Magnitude of this force: $F_{\text{on surface}} = f = \mu mg$ .
So, $\vec{F}_{\text{on surface}} = (\mu mg) \hat{i}$ .

3. Determine the Displacement of the Surface in Reference Frame 'S':

Reference frame 'S' moves with a constant velocity $\vec{v}_S = v_0 \hat{i}$ relative to the ground.
The surface is stationary in the ground frame, so its velocity relative to the ground is $\vec{v}_{\text{surface, G}} = 0$ .
The velocity of the surface relative to frame 'S' is given by:
$\vec{v}_{\text{surface, S}} = \vec{v}_{\text{surface, G}} - \vec{v}_{S, G} = 0 - v_0 \hat{i} = -v_0 \hat{i}$ .
So, in frame 'S', the surface moves with a constant velocity $-v_0 \hat{i}$ (i.e., to the left).

The time interval is from $t=0$ to $t=t_0$ .
The displacement of the surface in frame 'S' during this time interval is:
$\vec{s}_{\text{surface, S}} = \vec{v}_{\text{surface, S}} \times t_0 = (-v_0 \hat{i}) t_0$ .

4. Calculate the Work Done by Friction on the Surface in Frame 'S':

Work done $W = \vec{F} \cdot \vec{s}$ .
$W = \vec{F}_{\text{on surface}} \cdot \vec{s}_{\text{surface, S}}$
$W = (\mu mg \hat{i}) \cdot (-v_0 t_0 \hat{i})$
$W = - \mu mg v_0 t_0$

5. Substitute from Ground Frame Analysis:

From Equation 1, we have $v_0 = \mu g t_0$ . This implies $\mu g = v_0/t_0$ .
Substitute this into the work expression:
$W = - m (\mu g) v_0 t_0$
$W = - m (v_0/t_0) v_0 t_0$
$W = - m v_0^2$

The work done by the friction on the surface, as observed by reference frame 'S', is $-mv_0^2$ .

Explanation of the solution:

Determine the friction force exerted by the block on the surface. Since the block moves right, the friction force on the block is left. By Newton's third law, the force by the block on the surface is right, with magnitude $\mu mg$ .
Determine the velocity of the surface in the moving frame 'S'. Frame 'S' moves right with $v_0$ . The surface is stationary in the ground frame. Thus, in frame 'S', the surface moves left with velocity $v_0$ .
Calculate the displacement of the surface in frame 'S' over the time $t_0$ . Since the surface moves with constant velocity $-v_0$ (left), its displacement is $-v_0 t_0$ .
Calculate work done as the dot product of force and displacement. The force (right) and displacement (left) are opposite, so the work is negative: $W = (\mu mg)(-v_0 t_0)$ .
Relate $v_0$ , $\mu g$ , and $t_0$ from the block's motion in the ground frame. The block stops from $v_0$ due to acceleration $-\mu g$ in time $t_0$ , so $v_0 = \mu g t_0$ .
Substitute this relation into the work expression to get $W = -m v_0^2$ .