Work Power And Energy Ques 25
- Two blocks $A$ and $B$ are connected to each other by a string and a spring; the string passes over a frictionless pulley as shown in the figure. Block $B$ slides over the horizontal top surface of a stationary block $C$ and the block $A$ slides along the vertical side of $C$, both with the same uniform speed. The coefficient of friction between the surfaces of blocks is 0.2 . Force constant of the spring is $1960 N / m$. If mass of block $A$ is $2 kg$. Calculate the mass of block $B$ and the energy stored in the spring.
(1982, 5M)
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Answer:
Correct Answer: 25.$10 kg, 0.098 J$
Solution:
Formula:
- Normal reaction between blocks $A$ and $C$ will be zero. Therefore, there will be no friction between them.
Both $A$ and $B$ are moving with uniform speed. Therefore, net force on them should be zero.
For equilibrium of $A$
$$ \begin{aligned} m _A g & =k x \\ \therefore \quad x & =\frac{m _A g}{k}=\frac{(2)(9.8)}{1960} \\ & =0.01 m \end{aligned} $$
For equilibrium of $B$
$$ \begin{aligned} \mu m _B g=T & =k x=m _A g \\ \therefore \quad m _B & =\frac{m _A}{\mu}=\frac{2}{0.2} \\ & =10 kg \end{aligned} $$
Energy stored in spring
$$ \begin{aligned} U & =\frac{1}{2} k x^{2}=\frac{1}{2}(1960)(0.01)^{2} \\ & =0.098 J \end{aligned} $$
BETA
