Question

Two blocks of masses $0.25kg$ and $0.80kg$ are connected by means of a light inextensible string that goes over a smooth and massless pulley. If the work done (in joules) by string on the $0.25kg$ block during the first second of its motion after being released from rest is $\frac{998}{x}$ then find $x$ . (Take $g=10m{s}^{-2}$ )

Answer 1

Answer: 100

Acceleration is given by, $a=\frac{\text{ Net pulling force }}{\text{ Total mass }}=\frac{0.80g-0.25g}{0.80+0.25}=\frac{11g}{21}$
Displacement is given by, $s=\frac{1}{2}a{t}^2=\frac{1}{2}\left(\frac{11g}{21}\right){\left(1\right)}^2=\frac{11g}{42}$

Balancing force, we get,
$T-0.25g=0.25a$
$\therefore T=\frac{1}{4}g+\frac{1}{4}\left(\frac{11g}{21}\right)$
$\therefore T=\frac{8}{21}g$
Now, for $0.25kg$ mass,
$W_T=Tcos{0}^{\circ }=\frac{8}{21}g\times \frac{11}{42}g\times I=\frac{44}{441}\left(g^2\right)$
$\therefore W_T=9.98J$