A student pushes a baseball of m = 0.18 kg down onto the top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of d = 0.14 meters. The spring constant of the spring is k = 680 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring. m = 0.18 kgk = 680 N/md = 0.14 ma.) The ball is then released. What is its speed, v, in meters per second, just after the ball leaves the spring? b.) What is the maximum height, h, in meters, that the ball reaches above the equilibrium point? c.)  What is the ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point?

Question
Asked Nov 4, 2019

A student pushes a baseball of m = 0.18 kg down onto the top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of d = 0.14 meters. The spring constant of the spring is k = 680 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring.

m = 0.18 kg
k = 680 N/m
d = 0.14 m

a.) The ball is then released. What is its speed, v, in meters per second, just after the ball leaves the spring?

b.) What is the maximum height, h, in meters, that the ball reaches above the equilibrium point?

c.)  What is the ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point?

check_circleExpert Solution
Step 1

a)

The required speed from conservation of energy is,

1
1
2
2
680
(0.14)
0.18
-8.604 m/s
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Image Transcriptionclose

1 1 2 2 680 (0.14) 0.18 -8.604 m/s

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Step 2

b)

The required maximum...

1
-mgh
2
h=
2g
(8.604)
2(9.8)
-3.78 m
help_outline

Image Transcriptionclose

1 -mgh 2 h= 2g (8.604) 2(9.8) -3.78 m

fullscreen

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Mechanical Properties of Matter