Question

A remote controlled car of mass $m$ can move on a long plank of mass $M$ that rests on a horizontal floor. The mechanism of the car is so designed that it always moves on the plank with a constant velocity $v$ with respect to the plank. Coefficient of friction between the plank and the floor is $\mu$ and acceleration of free fall is $g$. How will velocity of the car with respect to the floor vary with time $t$ after it is switched on?

Answer 1

v

Answer 2

Let $v_{c,f}$ be the velocity of the car with respect to the floor, and $v_{p,f}$ be the velocity of the plank with respect to the floor. The velocity of the car with respect to the plank is given as $v_{c,p} = v$. The velocities are related by $v_{c,f} = v_{c,p} + v_{p,f}$. Assuming motion along a line, $v_{c,f} = v + v_{p,f}$.

Differentiating with respect to time, the accelerations are related by $a_{c,f} = \frac{dv}{dt} + a_{p,f}$. Since $v$ is constant, $\frac{dv}{dt} = 0$. Thus, $a_{c,f} = a_{p,f}$.

Let $a = a_{c,f} = a_{p,f}$.

The net external horizontal force on the car-plank system is the friction force on the plank from the floor, $f_k$.

By Newton's second law for the system, $f_k = (m+M)A_{CM}$, where $A_{CM}$ is the acceleration of the center of mass.

$A_{CM} = \frac{m a_{c,f} + M a_{p,f}}{m+M} = \frac{m a + M a}{m+M} = a$.

So, $f_k = (m+M)a$.

The kinetic friction force is $f_k = -\mu(M+m)g \frac{v_{p,f}}{|v_{p,f}|}$ for $v_{p,f} \ne 0$.

Thus, $(M+m)a = -\mu(M+m)g \frac{v_{p,f}}{|v_{p,f}|}$, which simplifies to $a = -\mu g \frac{v_{p,f}}{|v_{p,f}|}$.

Since $a = a_{p,f} = \frac{dv_{p,f}}{dt}$, we have $\frac{dv_{p,f}}{dt} = -\mu g \frac{v_{p,f}}{|v_{p,f}|}$ for $v_{p,f} \ne 0$.

Given the initial condition $v_{p,f}(0)=0$, this differential equation can only be satisfied for $t>0$ if $v_{p,f}(t)=0$. Any attempt for $v_{p,f}$ to become non-zero leads to a contradiction in the sign of acceleration and velocity.

Therefore, $v_{p,f}(t) = 0$ for all $t \ge 0$.

The velocity of the car with respect to the floor is $v_{c,f}(t) = v + v_{p,f}(t) = v + 0 = v$.

The velocity of the car with respect to the floor is a constant $v$.

The final answer is $\boxed{v}$.