Question

A block of mass $m(=0.1kg)$ 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 $2kg$ $m{s}^{-1}$ in $1s$ after the block starts from rest. Then, (given $g=10m{s}^{-2}$)

• 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 $F-mg=2$

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