NLM BYU 35 - Physics.lt

Solution 35 – Plank and Block

Calculation of Coefficient of Friction

1. Conservation of Momentum:
Initially, the plank ($M$) moves with velocity $u$, and the block ($m$) is at rest. Finally, both move together with velocity $v$. $$ Mu = (M+m)v \implies v = \frac{Mu}{M+m} $$

2. Work-Energy Theorem:
The work done by kinetic friction reduces the total kinetic energy of the system. The sliding distance is $l$. $$ \text{Work Done by Friction} = \Delta KE $$ $$ -f_k l = KE_f – KE_i $$ where $f_k = \mu m g$ is the frictional force.

$$ -\mu m g l = \frac{1}{2}(M+m)v^2 – \frac{1}{2}Mu^2 $$ Substitute $v = \frac{Mu}{M+m}$: $$ -\mu m g l = \frac{1}{2}(M+m)\left( \frac{Mu}{M+m} \right)^2 – \frac{1}{2}Mu^2 $$ $$ -\mu m g l = \frac{1}{2} \frac{M^2 u^2}{M+m} – \frac{1}{2}Mu^2 $$ $$ -\mu m g l = \frac{1}{2} M u^2 \left( \frac{M}{M+m} – 1 \right) $$ $$ -\mu m g l = \frac{1}{2} M u^2 \left( \frac{-m}{M+m} \right) $$ $$ \mu m g l = \frac{Mm u^2}{2(M+m)} $$

Solving for $\mu$:

$$ \mu = \frac{M u^2}{2gl(M+m)} $$

Alternate Solution: Work Done by Internal Forces

Friction between the block and the plank is an internal force for the (block + plank) system.

The net work done by a pair of internal forces is the same in every inertial frame. Hence, we may calculate the loss of kinetic energy in the centre-of-mass frame.

Initially, the relative velocity between the block and the plank is $$u_{\text{rel}} = u.$$

The kinetic energy associated with their relative motion is $$K_{\text{rel}} = \frac12 \left(\frac{mM}{m+M}\right)u^2,$$ where $$\frac{mM}{m+M}$$ is the reduced mass of the block-plank system.

When slipping stops, their relative velocity becomes zero. Therefore, the entire relative kinetic energy is dissipated by friction.

During slipping, the relative displacement between the two bodies is $l$. Hence the magnitude of work done by internal friction is $$W_{\text{friction}} = \mu m g\,l.$$

$$ \frac12\left(\frac{mM}{m+M}\right)u^2 = \mu m g l $$ Cancelling $m$, $$ \frac{Mu^2}{2(M+m)}=\mu gl $$ Therefore, $$ \boxed{\mu=\frac{Mu^2}{2gl(M+m)}} $$