A block of mass m = 5.50 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Let ha = 6.30 m.)An illustration shows a wavy track, starting from a crest, moving to a trough, then again to a crest and trough, and finally to a crest that then moves downward. Three points in the track are highlighted, A, B, and C. Point A is at the top of the track where an object of mass m is placed ready to get released. It is at the height labeled ha from the ground. Point B is shown at the next crest and is at a height of 3.20 meters from the ground. Point C is shown at the following trough and is at a height of 2.00 meters from the ground.(a) Determine the block's speed at points and .vB= m/svC= m/s(b) Determine the net work done by the gravitational force on the block as it moves from point to point .J

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Asked Nov 22, 2019
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A block of mass m = 5.50 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Let ha = 6.30 m.)

An illustration shows a wavy track, starting from a crest, moving to a trough, then again to a crest and trough, and finally to a crest that then moves downward. Three points in the track are highlighted, A, B, and C. Point A is at the top of the track where an object of mass m is placed ready to get released. It is at the height labeled ha from the ground. Point B is shown at the next crest and is at a height of 3.20 meters from the ground. Point C is shown at the following trough and is at a height of 2.00 meters from the ground.
(a) Determine the block's speed at points and .
vB = m/s
vC = m/s

(b) Determine the net work done by the gravitational force on the block as it moves from point to point .
J
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Expert Answer

Step 1

a)

From the conservation of energy at point A and B,

(98(63)-[9)(12)
7.79m/s
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(98(63)-[9)(12) 7.79m/s

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Step 2

Again, from conservation of en...

2g (A-A)
=2
(9.8)(63-2)
9.18m/s
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2g (A-A) =2 (9.8)(63-2) 9.18m/s

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