2. A bead of mass m = 8.35 kg is released from point A and slides...

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Question

Answer 1

Answer #1

2a) speed at point B, v = sqrt(2*KE/m)

= sqrt(2*mgh/m) where h is change in height

= sqrt(2*8.35*9.8*(6.2-3.2)/8.35)

= 7.668 m/s

b) Net work done by gravity = mgh where h is change in height

= 8.35*9.8*(6.20-2)

= 343.7 J

3a) work required to be done = mg sin theta *d

= 85*9.8*sin 42 degree*83

= 46263 J

3b] power = force*velocity

= mg sin theta *v

= 85*9.8*sin 42 degree*5.7

= 3177 W

Answer 2

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