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Question

74. In 2 minutes, a skilift raises 4 skiers at constant speed to a height of 140 m. The average mass of each skier is 65 kg. What is the average power provided by the tension in the cable pulling the lift?

76. A 63 kg skier coasts up a snow-covered hill that makes an angle of 25 with the horizontal. The initial speed of the skier is 6.6 m/s. after coasting a distance of 1.9 m up the slope, the speed of the skier is 4.4 m/s. a.) Find the work done by the kinetic frictional force that acts on the skis. b.) what is the magnitude of the kinetic frictional force?

Answer 1

74. Power P is the rate of energy flow and can be expressed as a product of applied force F and velocity V.

P=Fv

F= 4mg

V=h/t

P= 4(mg) h / t

P= 4(65 x 9.81) 140/(2 x 60)=2,980  Watts (or 3,000 W)

76. First let's set it up.

Work done if force F applied over a distance d.

W= Fd

Also

Ke= 0.5mV^2

Difference in Kinetic energy dKe

dKe=Ke1 - Ke2

dKe= 0.5m(V1^2 - V2^2)

dKe= W

Now we are ready to solve the problem

a) W=dKe= 0.5m(V1^2 - V2^2)

W= 0.5x63(6.6^2 - 4.4^2)

W= 760 Joules

However this how much energy was lost due to friction and gain in kinetic energy.

The skier  it will gain potential energy P=mgh

Pe= 63 x 9.81 x 1.9  sin(25)= 496 J ( or 500J)

finally work done by friction is then

Wf= W- Pe= 760 - 500 = 260 J

b) Then friction force f=Wf/d

f= 260/1.9 = 137 N ( 140 N)