Question

A block of mass ‘m’ slides down a rough inclined plane, as shown in figure. Friction coefficient between upper half of inclined plane and block is 0.2, while between lower half of inclined plane and block is 0.3. Length of inclined plane is 10 meter and angle of inclination ‘θ’ with horizontal is sin–1 (4/5). The velocity of block when it leaves the inclined plane will be

Answer 1

Answer:

Here, initial velocity , u=0

From the figure, net acceleration will be when it is sliding down,

frictional force=μmgcosθ

a=gsinθ−ugcosθ........................(1)

We know, Power = work done /time= rate of work done..................(2)

Also, Power =force *velocity.............(3)

Now, from equation of motion,

v=u+at

v=0+(gsinθ−ugcosθ)t (from eqn. (1)).....................(4)

From eqn. (2), (3) and (4),

work done /time= force *velocity,

=μmgcosθ.(gsinθ−ugcosθ)t

=μmg

2

tcosθ(sinθ−μcosθ)

Explanation:

hope it will help you.....