Search any question & find its solution
Question:
Answered & Verified by Expert
A block of mass $0.18 \mathrm{~kg}$ is attached to a spring of force constant $2 \mathrm{~N} / \mathrm{m}$. 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 $0.06 \mathrm{~m}$ and comes to rest for the first time. The initial velocity of the block in $\mathrm{m} / \mathrm{s}$ is $v=\frac{N}{10}$. Then, $N$ is
Solution:
2030 Upvotes
Verified Answer
The correct answer is:
4
Decrease in mechanical energy
$=$ Work done against friction
$\therefore \quad \frac{1}{2} m v^2-\frac{1}{2} k x^2=\mu m g x$
or $v=\sqrt{\frac{2 \mu m g x+k x^2}{m}}$
Substituting the values, we get
$$
v=0.4 \mathrm{~ms}^{-1}=\left(\frac{4}{10}\right) \mathrm{ms}^{-1}
$$
$\therefore$ Answer is 4 .
Analysis of Question
(i) Question is simple.
(ii) If $\mu_s$ and $\mu_k$ two values of coefficient of friction are given, then we will take $\mu_k$.
$=$ Work done against friction
$\therefore \quad \frac{1}{2} m v^2-\frac{1}{2} k x^2=\mu m g x$
or $v=\sqrt{\frac{2 \mu m g x+k x^2}{m}}$
Substituting the values, we get
$$
v=0.4 \mathrm{~ms}^{-1}=\left(\frac{4}{10}\right) \mathrm{ms}^{-1}
$$
$\therefore$ Answer is 4 .
Analysis of Question
(i) Question is simple.
(ii) If $\mu_s$ and $\mu_k$ two values of coefficient of friction are given, then we will take $\mu_k$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.