Work Power And Energy Ques 8
- A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 kg$ and $0.72 kg$. Taking $g=10 ms^{-2}$, find the work done (in Joule) by string on the block of mass $0.36 kg$ during the first second after the system is released from rest.
(2009)
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Answer:
Correct Answer: 8.8
Solution:
Formula:
- $a=\frac{\text { Net pulling force }}{\text { Total mass }}$
$$ =$\frac{0.72 g-0.36 g}{0.72+0.36}=\frac{0.36 g}{1.08}=\frac{g}{3}$ $$
$$ \begin{aligned} & \quad s=\frac{1}{2} a t^{2}=\frac{1}{2} \frac{g}{3}(1)^{2}=\frac{g}{6} \\ & T-0.36 g=0.36 a=0.36 \frac{g}{3} \quad 0.36 \text{ kg} \ & \therefore \quad T=0.48 g \\ & \text { Now, } \quad W _T=T S \cos 0^{\circ} \\ & \quad(\text { on } 0.36,\text{kg} \text { mass) } \ & =0.08\left(g^{2}\right)=0.08(10)^{2}=8\ \text{J} \end{aligned} $$
BETA
