Question

1. which statement best explains why running on a track with constant speed at 3 m/s is not work, but climbing a mountain at 1 m/s is work?

a. at constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy.
b. at constant speed, change in the potential energy is zero, but climbing a mountain produces change in the kinetic energy.
c. at constant speed, change in the kinetic energy is finite, but climbing a mountain produces no

Explanation:

To determine the correct answer, we analyze the concepts of work, kinetic energy (\(KE = \frac{1}{2}mv^2\)) and potential energy (\(PE = mgh\)):

1. Running on a track at constant speed:

• Kinetic energy depends on speed (\(v\)) and mass (\(m\)). Since speed is constant, \(KE\) does not change (\(\Delta KE = 0\)).
• For work to be done, there must be a change in energy (work - energy theorem: \(W = \Delta KE + \Delta PE\)). If \(\Delta KE = 0\) and the height (\(h\)) of the track is constant (so \(\Delta PE = 0\)), no net work is done.

1. Climbing a mountain:

• As height (\(h\)) increases, gravitational potential energy (\(PE = mgh\)) increases (\(\Delta PE > 0\)).
• Even if speed is constant (\(\Delta KE = 0\)), the change in potential energy means work is done (to overcome gravity and increase height).

Now, evaluate the options:

• Option a: Correctly states that constant speed means \(\Delta KE = 0\) (no work from KE change), while climbing a mountain increases \(h\), so \(\Delta PE > 0\) (work is done to change PE).
• Option b: Incorrect. Climbing a mountain at constant speed means \(\Delta KE = 0\) (speed is constant), not a change in KE.
• Option c: Incorrect. Constant speed means \(\Delta KE = 0\) (not "finite"), and climbing a mountain does produce a change in PE (so work is done).

Answer:

a. At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy.