Question

Two blocks of masses $0.25\,kg$ and $0.80\,kg$ 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 $0.25\,kg$ block during the first second of its motion after being released from rest is $\frac{998}{x}$ then find $x$ . (Take $g=10\,ms^{- 2}$ )

Answer 1

Acceleration is given by, $a=\frac{\text{ Net pulling force }}{\text{ Total mass }}=\frac{0 . 80\, g - 0 . 25 \,g}{0 . 80 + 0 . 25}=\frac{11 g}{21}$
Displacement is given by, $s=\frac{1}{2} at^{2}=\frac{1}{2}\left(\frac{11 g}{21}\right)\left(1\right)^{2}=\frac{11 g}{42}$

Balancing force, we get,
$T-0.25\,g=0.25\,a$
$\therefore T=\frac{1}{4}g+\frac{1}{4}\left(\frac{11 g}{21}\right)$
$\therefore T=\frac{8}{21}g$
Now, for $0.25\,kg$ mass,
$W_{T}=Tcos0^\circ =\frac{8}{21}g\times \frac{11}{42}g\times I=\frac{44}{441}\left(g^{2}\right)$
$\therefore W_{T}=9.98\,J$