In one of our snowy Chicago winter days, a bobsled run leads down a hill as sket

Question

In one of our snowy Chicago winter days, a bobsled run leads down a hill as sketched in the figure above. The bobsled starts from rest at point A and it's mass of the bobsled,including drive is 230 kg. All along the hills, between points A and D, friction is negligible, but at the end of the run between points D and E, the coefficient of kinetic friction is µk = 0.22.

1)

What is the speed of the bobsled at point B?

m/sec

2)

What is work done by gravity between A and C?

3)

Find the distance x beyond point D at which the bobsled will come to a halt.

Explanation / Answer

1)

Usin conservation of energy

kinetic energy at B = potential energy at A

(0.5) m Vb2 = mgha

Vb = sqrt(2 gha)

Vb = sqrt(2 x 9.8 x 50) = 31.3 m/s

b)

work done = change in PE at A and = mg(ha - hc) = 230 x 9.8 (50 - 30) = 45080 J

c)

KE at D = PE at A = mgha = 230 x 9.8 x 50 = 112700 J

frictional force , f = µk mg = 0.22 x 230 x 9.8 = 495.88 N

Work done by frictional force = KE at D

f x = 112700

(495.88) x = 112700

x = 227.3 m

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