A student pushes a baseball of m = 017 kg down onto the top of a vertical spring

Question

A student pushes a baseball of m = 017 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 = 0.18 meters. The spring constant of the spring is k = 650 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring. Randomized Variables m m = 0.17kg k = 650 N/m d = 0.18 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?

Explanation / Answer

Here ,

m = 0.17 Kg

d = 0.18 m

k = 650 N/m

a) let the speed of the ball is v

using conservation of energy

0.50 * 0.17 * v^2 = 0.50 * 650 * 0.18^2 - 0.17 * 9.8 * 0.18

solving for v

v = 11 m/s

the speed of ball is 11 m/s

b)

maximum height is h

0.50 * k * x^2 = m * g * h

0.50 * 650 * 0.18^2 = 9.8 * 0.17 * h

h = 6.321 m

the maximum height is 6.321 m

c)

at the half height

0.50 * m * v^2 = m * g * 6.321/2

v^2 = 61.94

v = 7.87 m/s

the speed of ball is 7.87 m/s

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