# A block of mass 3 kg starts from rest and slides down a surface, which corresponds to a quarter of a circle of 1.6 m radius. a) If the curved surface is smooth, calculate the speed of the block at the bottom. b) If the block’s speed at the bottom is 4 ms−1, what is the energy dissipated by friction as it slides down? c) After the block reaches the horizontal surface with a speed of 4 ms−1, it stops after travelling a distance of 3 m from the bottom. Find the frictional force acting on the horizontal surface due to the block.

1

2015-09-16T03:44:26+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
1)
change in KE = change in PE
1/2 m v² = m g R
v = √[2 g R] = √[2*10*1.6]  = √32 m/sec
============
2)
Loss of energy if the speed is only 4 m/s :  1/2 * 3 kg * [ (√32)² - 4² ]
= 24 Joules
This is the energy dissipated.

===========================
3)
Energy dissipated on the horizontal surface = 1/2 * 3 kg * 4² m²/s² = 24 Joules
work done by friction = Force * distance
24 J = Force * 3 m
Force = 8 N

click on thanks button above pls