Question

A woman of mass m = 77 kg runs at a velocity vi = 1.53 m/s before leaping on a skateboard that is initially at rest. After leaping on the board, the woman has a velocity vf = 1.45 m/s. What is the change in kinetic energy of the woman-skateboard system?

1) Cannot be determined
2) 0 J
3) 2.5 J
4) 5 J

Answer 1

The change in kinetic energy of the woman-skateboard system after the woman jumps onto the skateboard is approximately -8.9 J.

Step-by-step explanation:

To calculate the change in kinetic energy of the woman-skateboard system when a woman of mass m = 77 kg jumps from a running velocity of vi = 1.53 m/s onto a skateboard, resulting in a final velocity vf = 1.45 m/s, we use the following formula:

Change in kinetic energy = Final kinetic energy - Initial kinetic energy

Initial kinetic energy = (0.5) × m × (vi)2

Final kinetic energy = (0.5) × m × (vf)2

Substituting the given values:

Initial kinetic energy = (0.5) × 77 kg × (1.53 m/s)2

Final kinetic energy = (0.5) × 77 kg × (1.45 m/s)2

After computing the values:

Initial kinetic energy = 90.03675 J

Final kinetic energy = 81.15125 J

The change in kinetic energy = 81.15125 J - 90.03675 J

The change in kinetic energy = -8.8855 J

Therefore, the change in kinetic energy of the woman-skateboard system is approximately -8.9 J.