Work Power And Energy Ques 23
- A block of mass $0.18 kg$ is attached to a spring of force constant $2 N / m$. The coefficient of friction between the block and the floor is 0.1 . Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of $0.06 m$ and comes to rest for the first time. The initial velocity of the block in $m / s$ is $v=\frac{N}{10}$. Then $N$ is
(2011)
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Answer:
Correct Answer: 23.(4)
Solution:
Formula:
Modified Form of Work-Energy Theorem:
- Decrease in mechanical energy $=$ Work done against friction
$\therefore \frac{1}{2} m v^{2}-\frac{1}{2} k x^{2}=\mu m g x$ or $v=\sqrt{\frac{2 \mu m g x+k x^{2}}{m}}$
Substituting the values, we get
$$ v=0.4 m / s=\frac{4}{10} m / s \Rightarrow N=4 $$
BETA
