PartA A 25.0 kg child plays on swing having support ropes that are 2.40 m long.

Question

PartA A 25.0 kg child plays on swing having support ropes that are 2.40 m long. A friend pulls her back until the ropes are 41.0 froi the verlical and releases her from rest What is the potential energy for the child just as she is released, compared with the potential energy at tne bottom ot the swing? Part C How much work does the tension in the ropes do as the child swings from the in tial position to the bottom? Submit Part B How fast will shc be moving at the bottom of the swing? Submit Submit Request Answer

Explanation / Answer

Part A)

PE = mgh

We can find the height from the geometry of the triangle formed

With a 2.4 m long rope, when up at a 410 angle, it forms a triangle with sides of 2.4(cos41) = 1.81 m long.

Therefore the child will be 2.4 - 1.81 m above the lowest point of the swing, which is the height.

2.4 - 1.811 = .589 m

PE = (25)(9.8)(.589)

PE = 144.30 J

Part B)

For this part, use conservation of energy

KE = PE

.5mv2 = mgh (mass cancels)

v = [(2)(9.8)(.589)]

v = 3.39 m/s

Part C)

Apply the work energy theorem.

W = KE

W = .5(25)(02 - 3.392)

W = -143.6 J = 0

The negative is sign convention since the tension pulls upward as the child swings downward.

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