EXAMPLE 5.10Circus Acrobat An acrobat drops onto a springboard, causing it to co

Question

EXAMPLE 5.10Circus Acrobat

An acrobat drops onto a springboard, causing it to compress.

SOLUTION

Use conservation of mechanical energy.

(1)

(KE + PEg + PEs)i = (KE + PEg + PEs)f

The only nonzero terms are the initial gravitational potential energy and the final spring potential energy.

0 + mg(h + d) + 0 = 0 + 0 + ½kd2

mg(h + d) = ½kd2

Substitute the given quantities and rearrange the equation into standard quadratic form.

d2 - (0.123 m) d - 0.245 m2 = 0

Solve with the quadratic formula.

d = 0.560 m

LEARN MORE

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.)

Yes. The acrobat can provide mechanical energy by pushing herself up while in contact with the springboard.No. There is no external source of energy to provide the potential energy at a greater height.Yes. The acrobat can bend her knees while falling and then straighten them as if jumping when bouncing upward again.Yes. Elastic energy is always present in the spring and can give the acrobat greater height than initially.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.

PRACTICE IT

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

Explanation / Answer

Practice It:
mg(h+d) = 1/2kd2

45.4*9.8*(1.59+d) = 0.5*6.41*103*d2

707.4228+444.92d = 3205d2

d2-0.1388d-0.2207=0

d = [ 0.1388+sqrt(0.13882 - 4*1*(-0.2207)) ] / (2*1) = 0.5443 m

d =  0.5443 m

Exercise Hints:

using, mg(h+d) = 1/2kd2

7.79*9.8*(1.50+d) = 1/2*1.04*103*d2

114.5130+76.3420d = 520d2

d2 - 0.1468d - 0.2202 = 0

d = [ 0.1468+sqrt{0.14682 - (4*1*(-0.2202))} ] / (2*1) = 0.5484 m

d = 0.5484 m

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