Question

Skater with a mass of 50.0 kg slides across an icy

pond with negligible friction. As he approaches a
friend, both he and his friend hold out their hands,
and the friend exerts a force in the direction oppo-
site to the skater's movement, which slows the skat-
er's speed from 2.0 m/s to 1.0 m/s. What is the
change in the skater's kinetic energy?
A. –25 J
C. – 100J
B. -75J
D. -150J​

Answer 1

The change in the skater's kinetic energy is -75 J.

Step-by-step explanation:

The change in the skater's kinetic energy can be calculated using the work-energy theorem. The work done on an object is equal to the change in its kinetic energy. In this case, the skater's initial kinetic energy is given by 0.5 * mass * initial velocity^2, and the final kinetic energy is given by 0.5 * mass * final velocity^2. The change in kinetic energy is the difference between these two values.

The initial kinetic energy is 0.5 * 50.0 kg * (2.0 m/s)^2 = 100 J. The final kinetic energy is 0.5 * 50.0 kg * (1.0 m/s)^2 = 25 J. Therefore, the change in the skater's kinetic energy is 100 J - 25 J = 75 J.

So, the correct answer is B. -75 J.

Answer 2

B

Step-by-step explanation:

Calculate the kinetic energy before and after they hold hands. Then, to find the change in kinetic energy simply subtract the final KE by initial KE.