Question

For 7-9: A skydiver falls ot constant, terminal velocity in the presence of air resistance 7. (3 pts) The net work done on the skydiver is.. a) positive b) negative c) zero 8. (3 pts) The non-conservative work done on the skydiver is.. a) positive b) negative c) zero 9. (3 pts) The mechanical energy of the skydiver is. a) conserved c) decreasing b) increasing For each problem, show all work for full credit. Use at least three significant figures in your fino answer. Box or circle your final answer. Watch your units! 10. (14 pts) A physics student uses a spring launcher of spring constant 120 N/m to fire a 0.050 kg steel ball straight up in the air. Before being fired, the ball is at rest in the launcher and the spring is compressed 8 cm from its unstrained length. After being fired, the spring returns to its unstrained length and the ball reaches a maximum height 75 cm above its initial position. What work is done by non- conservative air resistance as the ball rises? Watch your units! hitting the bag, she is falling at 48.5 m/s straight down. She hits the bag and, after 1.2s of contact, i bounced straight up at 3 m/s. 11. (14 pts total)A 52 kg stuntwoman dives off a building. falling on to an air bag below. Just prior to forre (magnitude and direction) on the stunt woman during this collision.

Answer 1

Q.7. Using Newton's second law net force on the skydiver

$F_{net} = ma = m\times 0 = 0 N$ as object is moving with constant velocity so a = 0

So net work done on the on the skydiver will be

$W = F_{net}.h =0 N\times h = 0 J$

So option C is correct.

Q.8. Work done by the non conservative force ( air resistance / drag force) will be negative becuase drag force act in the opposite direction of the vertical displacement

$W = F_{d}.h\cos (180^{0}) = - F_{d}h$

So B is correct.

Q.9. Decreasing because

As KE remains constant but PE is decreasing so total ME is decreasing.

Q.10. Initial elastic potential energy

$PE_{i} = \frac{1}{2}kx^{2}= \frac{1}{2}\times 120 N/m \times (8\times 10^{-2}m)^{2} = 0.384 J$

Final gravitational PE

$PE_{f} = mgh = 0.050 kg\times 9.8 m/s^{2}\times 75\times 10^{-2}m = 0.3675 J$

Work done by non conservative force is given by using work energy theorem