Question

Sarah is holding a 20 kg dumbbell over a 30-meter high cliff (32 meters from her hands to the ground). Just as she releases the dumbbell, it has a potential energy of 6272 J and kinetic energy of 0 J. However, just as the dumbbell hits the ground, it has a potential energy of 0 J and kinetic energy of 6272 J. How much work was done on the dumbbell from the second it leaves Sarah's hand to the second it hits the ground?

Answer 1

Work-Energy Theorem states that the work done by the sum of all forces acting on an object equals the change in the kinetic energy of the object.

Therefore, work done on the dumbbell = Final Kinetic Energy - Initial Kinetic Energy

= 6272 J - 0 J

= 6272 J

Alternatively,

Here, the force acting on the dumbbell, F = weight = mg = 20 * 9.8 N = 196 N

Total displacement of the dumbbell while falling to the ground, S = 32 m

Therefore, total work done on the dumbbell, W = Force * Displacement = F * S = 196 * 32 = 6272 J