Question

An ice cube of mass $M_0$ is given a velocity $v_0$ on a rough horizontal surface with coefficient of friction $\mu$. The block is at its melting point and latent heat of fusion of ice is $L$. The block receives heat only due to the friction forces and all work is converted into heat. Find the mass ($M_t$) of the remaining ice block after time $t$.

• A. $M_t=M_0{e}^{-\frac{2\mu g}{L}(v_{0}t+\frac{1}{2}\mu g{t}^2)}$
• B. $M_t=M_0{e}^{-\frac{\mu g}{L}(v_{0}t-\frac{1}{2}\mu g{t}^2)}$
• C. $M_t=M_0{e}^{-\frac{3\mu }{L}(v_{0}t-\frac{1}{2}\mu g{t}^2)}$
• D. $M_t=M_0{e}^{-\frac{2\mu g}{L}(v_{0}t-\mu g{t}^2)}$

Answer 1

The correct option is B $M_t=M_0{e}^{-\frac{\mu g}{L}(v_{0}t-\frac{1}{2}\mu g{t}^2)}$

Given that,
Initial mass $=M_0$
Initial velocity $=v_0$
Coefficient of friction $=\mu$
Latent heat of fusion of ice $=L$
For any instant, let its mass be $m$ and velocity be $v$.
Frictional force at an instant $=f=\mu N=\mu mg$
Heat generated due to friction per sec $=P=f.v=\mu mgv$
Let the rate of melting of ice be $\frac{dm}{dt}$
So, heat required per sec $=-\left(\frac{dm}{dt}\right)L$
Also, $\mu mgv=-\left(\frac{dm}{dt}\right)L$ {minus sign indicates decrease of mass with time}
$\frac{dm}{m}=\frac{-\mu gv}{L}dt\dots \dots \left(i\right)$

As $vdt=ds,$ where $ds$ is displacemnt of block in $dt$ time interval.
$\frac{dm}{m}=\frac{-\mu g}{L}ds\dots \dots \left(i\right)$
$\Rightarrow \underset{M_0}{\overset{M_t}{\int }}\frac{dm}{m}=-\frac{\mu g}{L}\underset{0}{\overset{S}{\int }}ds$
$\Rightarrow \text{ln}\left(\frac{M_t}{M_0}\right)=-\frac{\mu g}{L}S$
$\Rightarrow M_t=M_0{e}^{-\frac{\mu g}{L}S}$
As initial velocity is $v_0$ and retardation due to friction is $\mu g,$ we get
$S=v_{0}t-\frac{\mu g{t}^2}{2}$
Substituting in above equation, we get


$\Rightarrow M_t=M_0{e}^{-\frac{\mu g}{L}(v_{0}t-\frac{\mu g{t}^2}{2})}$