A 54.5-kg skateboarder starts out with a speed of 1.85 m/s. He does +80.0 J of w

Question

A 54.5-kg skateboarder starts out with a speed of 1.85 m/s. He does +80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 6.20 m/s.

(a) Calculate the change (PE = PEf - PE0) in the gravitational potential energy.


(b) How much has the vertical height of the skater changed?
m
Is the skater above or below the starting point?

below the starting pointabove the starting point

Explanation / Answer

(a) The kinetic energy change is
K = 1/2 m (vf² - vo²) = 1/2 x 54.5 x (6.2² - 1.85²) = 954.23 J
According to the kinetic energy theorem, K equals the work done bay all the forces (conservative and not conservative) acting on the system.
Therefore
K = 80 - 265 - (PE)
whence
(PE) = 80 - 265 - 954.23 = - 1139.23 J

(b) The height change is
h = (PE) / m g = 1139.23 / (54.5x 9.81) = 2.13m (in absolute value)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.