A Classical Approach of Newtonian Mechanics

5 CONSERVATION OF ENERGY 5.8 Power

B

.,

By energy conservation, ∆K = −∆U, where K represents kinetic energy. How-
ever, since the initial kinetic energy is zero, the change in kinetic energy ∆K is

equivalent to the final kinetic energy KB. Thus,

KB = −∆U = 5.30 × 104 J.

Now, KB = (1/2) m v 2 , where vB is the final speed. Hence,

vB =

‚

., 2 KB

‚ 2 × 5.3 0 × 104

300

= 18.8 m/s.

Worked example 5.5: Sliding down a plane

Question: A block of mass m = 3 kg starts at rest at a height of h = 43 cm on a

plane that has an angle of inclination of θ = 35 ◦ with respect to the horizontal.

The block slides down the plane, and, upon reaching the bottom, then slides

along a horizontal surface. The coefficient of kinetic friction of the block on both

surfaces is μ = 0.25. How far does the block slide along the horizontal surface

before coming to rest?

Answer: The normal reaction of the plane to the block’s weight is

R = m g cos θ.

Hence, the frictional force acting on the block when it is sliding down the plane

is

f = μ R = 0.25 × 3 × 9.81 × cos 35 ◦ = 6.03 N.

The change in gravitational potential energy of the block as it slides down the

plane is

∆U = −m g h = − 3 × 9.81 × 0.43 = −12.65 J.

The work W done on the block by the frictional force during this process is

W = −f x,

m

=