LEARN MORE REMARKS The other solution, d = -0.437 m, can be rejected because d w

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REMARKS The other solution, d = -0.437 m, can be rejected because d was chosen to be a positive number at the outset. A change in the acrobat's center of mass, say, by crouching as she makes contact with the springboard, also affects the spring's compression, but that effect was neglected. Shock absorbers often involve springs, and this example illustrates how they work. The spring action of a shock absorber turns a dangerous jolt into a smooth deceleration, as excess kinetic energy is converted to spring potential energy.

QUESTION Is it possible for the acrobat to rebound to a height greater than her initial height? Explain. (Select all that apply.)

1. No. There is no external source of energy to provide the potential energy at a greater height.

2. No. The kinetic energy that the acrobat gains on the way down is converted entirely back into potential energy when she reaches the initial height.

3. Yes. The acrobat can bend her knees while falling and then straighten them as if jumping when bouncing upward again.

4. Yes. Elastic energy is always present in the spring and can give the acrobat greater height than initially.

5. Yes. The acrobat can provide mechanical energy by pushing herself up while in contact with the springboard.

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Explanation / Answer

m = mass of acrobat = 48.7 kg

h = height above the spring board = 2.50 m

d = compression of the spring

k = spring constant = 8.49 x 103 N/m

Using conservation of energy

Potential energy of acrobat at the top = spring potential energy

mg(h + d) = (0.5) k d2

48.7 x 9.8 (2.5 + d) = (0.5) (8.49 x 103 ) d2

d = 0.589 m

#

m = mass of acrobat = 8.47 kg

h = height above the spring board = 1.06 m

d = compression of the spring

k = spring constant = 1.19 x 103 N/m

Using conservation of energy

Potential energy of acrobat at the top = spring potential energy

mg(h + d) = (0.5) k d2

8.47 x 9.8 (1.06 + d) = (0.5) (1.19 x 103 ) d2

d = 0.461 m

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