A slingshot obeying Hooke\'s law is used to launch pebbles vertically into the a

Question

A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 23.3cm against the elastic band, the pebble goes 6.1m high.

Part A: Assuming that air drag is negligible, how high will the pebble go if you pull it back 46.6cm instead?Express your answer to two significant figures and include the appropriate units.

Part B: How far must you pull it back so it will reach 12.2m? Express your answer to three significant figures and include the appropriate units

Part C: If you pull a pebble that is twice as heavy back 23.3cm , how high will it go? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

We need to apply conservation of energy

Part A)

Elastic PE = Gravitational PE

.5kx2 = mgh

Thus k/m = 2gh/x2

Set up the ratio

2gh/x2 = 2gh/x2 (2g cancels)

h/x2 = h/x2

6.1/(.233)2 = h/(.466)2

h = 24 m

Part B)

Apply the same ratio...

12.2/x2 = 6.1/(.233)2

x = .330 m (33.0 cm)

Part C)

Since we start with...

.5kx2 = mgh

k = 2mgh/x2

2mgh/x2 = 2mgh/x2 (2 and g cancel.)

mh/x2 = 2mh/x2

h = 2h

6.1 = 2h

h = 3.1 m

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