Question

A $2kg$ of block of wood rests on a long table top. A $5g$ bullet moving horizontally with a speed of $150m/s$ is shot into the block and sticks to it. The block then slides $2.7m$ along the table top and comes to a stop. The force of friction between the block and the table is:

• A. $0.052N$
• B. $3.63N$
• C. $2.50N$
• D. $1.04N$

Answer 1

The correct option is A $0.052N$

$Given$ :- Mass of block $\left(m\right)$ = 2 Kg

Mass of bullet $\left(m_1\right)$ = 5 g = 0.005 kg

Speed of bullet $\left(v_1\right)$ = 150 m/s

Distance slide by block $\left(s\right)$ = 2.7m

$ToFind$ :- frictional force $\left(f\right)$ between block and table

$Solution$ :- Let the Speed of sliding of block on table top be $v_2$

Since we know , By linear momentum i.e,

$m_1{v}_1=m_2{v}_2$

$⟹$ (0.005) × 150 = (2.005) × $v_2$

$\therefore$ $v_2$ = 0.374 m/s

Now , By work energy theorem we get ,

$f\times s=\frac{1}{2}(m+m_1)(v_2{)}^2$

$⟹$ $f\times \left(2.7\right)=\frac{1}{2}\left(2.005\right)(0.374{)}^2$

$\therefore$ $\underset{–}{f=0.052N}$

Hence , $\underset{–}{OptionA\left(0.052N\right)iscorrect.}$