A block of mass 2.0 kg lies on a rough…

A block of mass 2.0 kg lies on a rough horizontal table and is attached to a horizontal spring of force constant 200 N m^-1, the other end being…

Numerical answer

A block of mass

2.0\ \mathrm{kg}

lies on a rough horizontal table and is attached to a horizontal spring of force constant

200\ \mathrm{N\,m^{-1}}

, the other end being fixed to a wall. The coefficient of friction (static and kinetic) between block and table is

0.10

and

g=10\ \mathrm{m\,s^{-2}}

. The block is pulled so that the spring is stretched by

0.305\ \mathrm{m}

from its natural length and released from rest. Find the total distance (in metres) travelled by the block before it finally comes to rest. Give the answer correct to two decimal places.

Show answer and solution

Answer: 4.65

Friction force

f=\mu mg=2.0\ \mathrm{N}

. The block can remain at rest wherever

kx\le f

, i.e.

|x|\le f/k=0.01\ \mathrm{m}

. In each half-cycle the amplitude decreases by

2f/k=0.02\ \mathrm{m}

, so the successive extreme positions are

0.305,\ 0.285,\ 0.265,\dots,\ 0.025,\ 0.005\ \mathrm{m}

(alternating sides). After the 14th half-cycle the block sits at

0.025\ \mathrm{m}

, where

kx=5\ \mathrm{N}>f

, so it moves once more, to

0.005\ \mathrm{m}

on the other side, where

kx=1\ \mathrm{N}<f

and it stays. Apply the work-energy theorem over the whole motion:

\tfrac12k(x_0^2-x_f^2)=f\,s

, so

s=\dfrac{100\,(0.305^2-0.005^2)}{2}=\dfrac{100\times0.093}{2}=4.65\ \mathrm{m}

Report a mistake

Next Work, Energy and Power question →

Original question written in the JEE Advanced 2025 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake