A 70-kg runner begins his slide into second base where he is moving at a speed of 40.0 m/s. The coefficient of kinetic friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.

(a) How much mechanical energy is lost due to friction acting on the runner.

(b) How far does he slide?

Given Information:

Initial speed = vi = 40 m/s

Final speed = vf = 0 m/s

Mass of runner = m = 70 kg

Coefficient of friction = k = 0.70

Required Information:

a) Energy lost = ?

b) Distance = ?

Answer:

a) Energy lost = 56000 Joules

b) Distance = 116.6 meters

Explanation:

As we know the kinetic energy is given by

KE = 0.5mvi²

KE = 0.5*70 * (40) ²

KE = 56000 J

The work done by the runner is given by

W = Fd

We also know that the work is done in the form of KE

KE = Fd

where F = mgk substitute in the above equation so equation becomes

KE = mgkd

d = KE/mgk

where g = 9.8 m/s² is acceleration due to gravity and and k = 0.70 is friction coefficient.

d = 56000/70*9.8*0.70

d = 116.6 m

Therefore, the runner slide for 116.5 m after he stopped at the base.

(a) K. E = 56000 J = 56 KJ, (b) d = 116.618 m

Explanation:

Given:

m = 70 Kg, Vi = 40.0 m/s, Vf = 0 m/s, μk = 0.70

Solution:

(a) K. E. = ? J (due to motion of the runner the mechanical energy loss is in the form K. E.)

K. E. = 1/2 m v² = 0.5 * 70 kg * (40.0 m/s) ²

K. E = 56000 J = 56 KJ

(b) distance d = ? m

W = F * d

∴W = K. E = 56000 J and F = mg μk

K. E. = mg μk * d

so 56000 J = 70 kg * 9.8 m/s² * 0.70 * d

d = 116.618 m