Question

A 2.0-kg ball is dropped from a height of 5.0 meters. It hits the ground and bounces upward. If it reaches a height of only 3.2 m, determine the magnitude of the work done by non-conservative forces in this situation.

Answer #1

From work energy theorem,

Work done by non conservative forces(W) = change in kinetic energy(dKE)

W = KEf - KEi

W = 0.5m*Vf^2 - 0.5m*Vi^2

here, m = mass = 2.0 kg

Vf = speed after bounce = ??

from third kinematics law,

v^2 - u^2 = 2*g*h

here, h = height = 3.2 m

v = final speed at maximum height = 0

So, u = initial speed = Vf = sqrt(2*g*h) = sqrt(2*9.81*3.2)

Vf = 7.9236 m/s

now, Vi = initial speed before hitting the ground = ??

Similarly, from third kinematics law,

v^2 - u^2 = 2*g*h

here, h = height = 5.0 m

v = final speed at ground = Vi = ?

u = 0 (dropped)

So, v = initial speed = Vi = sqrt(2*g*h) = sqrt(2*9.81*5.0)

Vi = 9.9045 m/s

then, W = 0.5*2.0*7.9236^2 - 0.5*2.0*9.9045^2

W = -35.3 J

Therefore magnitude of Work done by non conservative forces =
**|W| = 35.3 J**

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