Question

A truck carries a $2$ ton stone-block on a level horizontal road. The coefficient of static and kinetic frictions between the stone-block and truck-bed are $0.4$ and $0.2$, respectively. When the truck accelerates at $2\frac{m}{s^2}$, how much friction (in $N$) acts between the stone-block and the truck-bed? $g=10\frac{m}{s^2}$

Answer 1

Given: $m=2ton=2000kg,\mu =0.4$ between stone and truck
Below is $FBD$ of stone block on the truck

Friction is responsible for providing
acceleration to the block
Maximum acceleration friction can provide is
$\mu g=0.4g=4\frac{m}{s^2}$
Here truck moves with $2\frac{m}{s^2}$
Hence block will also move with $2\frac{m}{s^2}$
$\therefore f=ma=2000\times 2=4000N$

Final answer: 4000 N