JEE Main 2023 · Physics · Kinematics and Laws of Motion · medium

A block of mass 2 kg is projected up a rough incline of inclination θ with sin θ=0.6, with an initial speed of 10 m s^-1. The coefficient of kinetic…

Numerical answer

A block of mass

2\ \mathrm{kg}

is projected up a rough incline of inclination

\theta

with

\sin\theta=0.6

, with an initial speed of

10\ \mathrm{m\,s^{-1}}

. The coefficient of kinetic friction is

0.5

and the incline is long enough for the block to stop and slide back. Its kinetic energy (in J) when it returns to the starting point is (

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

)

Show answer and solution

Answer: 20

Going up:

a_1=g(\sin\theta+\mu\cos\theta)=10(0.6+0.4)=10\ \mathrm{m\,s^{-2}}

, so it travels

s=\dfrac{10^2}{2\times10}=5\ \mathrm{m}

. Coming down (the block slides since

\tan\theta>\mu

a_2=g(\sin\theta-\mu\cos\theta)=10(0.6-0.4)=2\ \mathrm{m\,s^{-2}}

. Then

v^2=2a_2s=20

and

K=\tfrac12(2)(20)=20\ \mathrm{J}

. Check by work-energy: friction work over the round trip

=2\mu mg\cos\theta\,s=2(0.5)(20)(0.8)(5)=80\ \mathrm J

, and

100-80=20\ \mathrm J

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