TS EAMCET 6-Aug-2021 Shift 2 Question Paper

The disc is at rest at the top of the hill.
∴ Potential energy of the disc, U=mgh
Let the velocity of the sliding disc before the friction is v.
∴‌ The kinetic energy, ‌K=‌

1

2

mv2
m=1g From energy conservation,
‌

1

2

mv2‌‌=mgh
v‌‌=√2gh
The disc now gets onto the plank and slows down and moves as one piece with the plank.
Let the velocity of the plank is v′.
From conservation of momentum,
mv‌‌=(m+M)v′
⇒‌‌v′‌‌=‌

m√2gh

(m+M)

∴ The kinetic energy of the plank,
‌K2=‌

1

2

(m+M)v2=‌

m2×2gh

2(M+m)

⇒‌‌K2‌‌=‌

m2gh

(m+M)

∴ The total work done by the frictional force is Wf=K1−K2=mgh−‌

m2gh

m+M

=‌

Mmgh

m+M

⇒‌‌Wf=‌

100×10−3×1×10−3×10×10

(100+1)×10−3

⇒‌‌Wf=0.1J