# Terminal velocity of pilot; ideal pulley; pop gun's spring

1. In 1955 a pilot fell 370 meters without his parachute opening. He landed in a snowbank, making a crater 1.1 meters deep. He walked away with minor injuries. If his mass was 80 kg and his terminal velocity was 35 m/s, calculate

a) the work done on him by the snow

b) the average force exerted on him by the snow

c) the work done on him by air resistance

2. An ideal pulley has a string over it, with masses attached to either end. One end has a 5kg mass, sitting on the floor. The other end has an 8kg mass, which starts from rest but the string is so short that it starts 3 meters above the floor. Use the work-energy theorem to calculate the masses' speeds when the 8kg mass reaches the floor

3. My 'pop' gun is essentially a spring that fires marbles as projectiles. If I load the gun by compressing the spring 9cm with 7 newtons of force, how fast will it project a 25 gram marble?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Problem 1.

The physics you need to apply to this problem is that energy is changed from one form to another. If potential energy PE is lost then energy appears in some form(s) in equal quantity.

Let M= 80 kg, H= 370 m, D= 1.1 m, and v= 35 m/s.

Part a1. In moving distance D into the snow in coming to rest: Work done by snow = PE lost in snowbank + KE lost in stopping

W= M g D + .5 M v^2 = (80 kg)(9.8 nt/kg)(1.1 m) + (.5)(80 kg)(35^2 m^2/sec^2) from which W= 49862 nt m or Joules

Part b. The average force by the snow, Fsnow times D must equal KE lost + PE lost in moving distance D

Fsnow D = .5 M v^2 + M g D from which (Fsnow)(1.1 m) = (.5)(80 kg)(35^2 m^2/sec^2) + (80 kg)(9.8 nt/kg)(1.1 m) from which Fsnow = 45329 nt

Part c. For constant velocity, the force exerted by the air must equal the force of gravity so the net force is zero. This gives:

Fair = M g = (80 kg)(9.8 nt/kg) = 784 nt.

Problem 2.

In this event, M1= 5 kg and M2= 8 kg and Height lost by M2 and gained by M1 = H= 3m, final speed v of each.

PE lost by M2 = KE gain by M1 + KE gain by M2 + PE gain by M1

M2 g H = .5 M1 v^2 + .5 M2 v^2 + M1 g H from which (8 kg)(9.8 nt/kg)(3 m)=(5 kg)(9.8 nt/kg)(3 m) + .5 (13 kg)(v^2)

from which: v= 3.68 m/sec

Problem 3.

The force F= 7 nt on the spring when compressed distance d= .09 m give spring's force constant k=F/d=7nt/.09m

from which k= 78 nt/m. Energy given up by spring upon release equals kinetic energy of marble.

.5 k d^2 = .5 M v^2 from which v^2=(d^2) (k/M)=(.09^2) (78 nt/m)/(.025 kg)

from which v= 5.03 m/sec

The conservation of energy is extremely important in physics.

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