Answer :

Mass of the block, m = 200 g = 0.2 kg

Acceleration due to gravity, g = 10 m/s2


Length of the incline plane, L = 10 m


Height of the incline plane, h = 3.2 m


(a.) Required work for lifting the block to the top of the incline = mgh


= 0.2×10×3.2


= 6.4 J


(b.) Work done against friction = 0 J (as the surface is frictionless)


Required work for sliding the block up = Work done against gravity


= 6.4 J


(c.) Let the speed of the block when it falls off the incline be v m/s.


The speed of the block when it is at rest on the top, u = 0 m/s


According to equation of motion, v2 - u2 = 2gh


v2 - 0 = 2×10×3.2


v2 = 64


v = 8 m/s


(d.) When the block slides down, let the speed of the block be v m/s


Change in kinetic energy of the block = work done by gravity + work done by friction


1/2 mv2 = mgh + 0


v2 = 2gh


v2 = 2×10×3.2 = 64


v = 8 m/s


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

A rectangular boxPhysics - Exemplar

A helicopter of mPhysics - Exemplar

There are four foPhysics - Exemplar

When a body <spanPhysics - Exemplar

Block <span lang=Physics - Exemplar

A block <span lanPhysics - Exemplar

Two <span lang="EPhysics - Exemplar