You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and

Question

You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it. Imagine that you slowly pull down with a spring scale so the block is now below the equilibrium position where it was hanging at rest. The scale reading when you let go of the block is 3.0 N.

A) Where was the block when you let go? Assume y0 is the equilibrium position of the block and that "down" is the positive direction.

B) Determine the work you did stretching the spring.

C) What was the energy of the spring-Earth system when you let go (assume that zero potential energy corresponds to the equilibrium position of the block)?

D)How far will the block rise after you release it?

E)About 90% of vibrational energy are lost in 50 cycles. Qualitatively represent the beginning of cycle #1 and beginning of cycle # 25 with an energy bar chart.

Explanation / Answer

Ans:- Given data

Yi = 0.07m,   m=0.1kg, F = 3N

A] Spring force law F = - k*x

F= m*a = - k* x

   = 0.1* 9.8 = -k*0.07

K = 14 N/m

F = - 14* Y

-3 = -14*Y

Y = 0.21m

B] work done by spring

Ws = ½*14*(0.07)2 – ½*14*(0.21)2

Ws = - 0.2744J

C] U = ½*k*Yf2 – ½*k*Yi2

zero potential energy corresponds to the equilibrium position

U = ½*k*Yf

    = ½ * 14*(0.21)2

    = 0.3087J

The block rise after realse

Ym = 0.21m

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