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yawa3891 [41]
3 years ago
13

A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T

he block sticks to the spring and momentarily stops after compressing the spring 14.9 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 3h0 above the spring, what would be the maximum compression of the spring?
Physics
1 answer:
Nina [5.8K]3 years ago
4 0

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= \frac{1}{2}kx² = \frac{1}{2}(530)(0.149)² =  5.88 J

b)  Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mgh_o = \frac{1}{2}kx² - mgx

And, h_o= ( \frac{1}{2}kx² - mgx )/(mg) = [\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)]/(0.645x9.8)    

   h_o=   0.78m            

d) Now, if the initial initial height of block is 3h_o

h_o = 3 x 0.78 = 2.34m

then, \frac{1}{2}kx² - mgx - mgh_o =0

 

\frac{1}{2}(530)x²  - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0  

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum  compression of the spring)

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Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

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        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

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The force needed to keep the space shuttle moving at constant speed is 0.

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