Question

A block of mass $m (= 0.1 kg)$ is hanging over a frictionless light fixed pulley by an inextensible string of negligible mass. The other end of the string is pulled by a constant force $F$ in the verticaly downward direction. The linear momentum of the block increase by $2\,kg$ $ms^{-1}$ in $1\,s$ after the block starts from rest. Then, (given $g \,= \,10\,ms^{-2}$)

• A. The tension in the string is F
• B. The tension in the string is 3N
• C. The work done by the tension on the block is 20 J during this 1 s
• D. The work done against the force of gravity is 10 J

Answer 1

Answer: (A), (B), (D)

The free body diagram $F-m g=2$

$\Rightarrow F=2+m g=3 N$
also, $a=\frac{\text { unbalanced force }}{\text { mass }}=\frac{2}{0.1}=20 \,m / s ^{2}$
$\therefore S=\frac{1}{2} a t^{2}=\frac{1}{2} \times 20 \times 1=10 \,m$
Hence, work done by tension
$=F \times 10=3 \times 10=30\,J$
So, the work done against gravity
$=m g \times S=1 \times 10=10 \,J$