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Question: A very large plank $P$ of some unknown mass is being moved with velocity $v_0\hat{i}$ under applicat...

A very large plank PP of some unknown mass is being moved with velocity v0i^v_0\hat{i} under application of an external force (not shown in figure). Simultaneously a block BB of mass mm placed on the plank is also moving with velocity v0k^v_0\hat{k}. All these velocities are with respect to ground frame and at t=0t=0. Coefficient of friction between the plank and the block μ\mu. Choose the correct option(s).

A

Kinetic friction force acting on the block at t=0t=0 is −μmgk^-\mu mg\hat{k}

B

At t=0t=0, power developed (with respect to ground frame) by kinetic friction force on the block B is −μmgv02\frac{-\mu mgv_0}{\sqrt{2}}.

C

At t=0t=0, power developed (with respect to ground frame) by kinetic friction force on the block B is −μmg2v0-\mu mg\sqrt{2}v_0.

D

At t=0t=0, heat dissipation per second in the system is 22μmgv02\sqrt{2}\mu mgv_0.

Answer

Option B is correct

Explanation

Solution

  1. Relative Velocity:

The plank has velocity v⃗P=v0i^\vec{v}_P = v_0 \hat{i} and the block has v⃗B=v0k^\vec{v}_B = v_0 \hat{k}.
Thus, the block’s velocity relative to the plank is

v⃗B/P=v⃗B−v⃗P=v0(k^−i^)\vec{v}_{B/P} = \vec{v}_B - \vec{v}_P = v_0(\hat{k}-\hat{i})

with magnitude ∣v⃗B/P∣=v02+v02=2 v0|\vec{v}_{B/P}| = \sqrt{v_0^2+v_0^2}=\sqrt{2}\,v_0.

  1. Friction Force on the Block:

Kinetic friction opposes the relative motion. Its magnitude is μmg\mu mg. Therefore, the friction force on the block is

f⃗B=−μmg v⃗B/P∣v⃗B/P∣=−μmg k^−i^2.\vec{f}_B = -\mu mg\,\frac{\vec{v}_{B/P}}{|\vec{v}_{B/P}|} = -\mu mg\,\frac{\hat{k}-\hat{i}}{\sqrt{2}}.

Hence, it is not solely in the k^\hat{k} direction. (Option A is false.)

  1. Power Developed by Kinetic Friction on the Block:

Power P=f⃗B⋅v⃗BP = \vec{f}_B \cdot \vec{v}_B with v⃗B=v0k^\vec{v}_B = v_0\hat{k}:

P=−μmg k^−i^2⋅v0k^=−μmg v0 (1−0)2=−μmgv02.P = -\mu mg\,\frac{\hat{k}-\hat{i}}{\sqrt{2}} \cdot v_0\hat{k} = -\mu mg\,\frac{v_0\,(1-0)}{\sqrt{2}} = -\frac{\mu mgv_0}{\sqrt{2}}.

(Option B is correct; Option C is false.)

  1. Heat Dissipation (Rate):

The heat dissipated is the sum of the work rates done by friction on both bodies. The plank experiences the equal and opposite friction force f⃗P=−f⃗B\vec{f}_P = -\vec{f}_B and moves with velocity v0i^v_0\hat{i}:

PP=f⃗P⋅v⃗P=μmg k^−i^2⋅v0i^=−μmgv02.P_P = \vec{f}_P \cdot \vec{v}_P = \mu mg\,\frac{\hat{k}-\hat{i}}{\sqrt{2}} \cdot v_0\hat{i} = -\frac{\mu mg v_0}{\sqrt{2}}.

Adding the block's contribution:

Ptotal=−μmgv02−μmgv02=−2 μmgv0.P_{\text{total}} = -\frac{\mu mgv_0}{\sqrt{2}} -\frac{\mu mgv_0}{\sqrt{2}} = -\sqrt{2}\,\mu mgv_0.

The magnitude of heat dissipation is 2 μmgv0\sqrt{2}\,\mu mgv_0 (not 22μmgv02\sqrt{2}\mu mgv_0); (Option D is false).