Question

Problem 4: A student pushes a baseball of m= 0.14 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= 0.16 meters. The spring constant of the spring is k = 770 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring. Randomized Variables m= 0.14 kg k= 770 N/m d= 0.16 m Part (a) The ball is then released. What is its speed, v, in meters per second, just after the ball leaves the spring? Numeric A numeric value is expected and not an expression V= Part (b) What is the maximum height, h, in meters, that the ball reaches above the equilibrium point? Numeric Anumeric value is expected and not an expression. h= Part (c) What is the ball's velocity, in meters per second, at half of the maximum height relative to the equilibrium point? Numeric Anumeric value is expected and not an expression. 10.5 =

General Physics

Answer 1

• Mass of the baseball:m = 0.14 kg

• Spring constant:k = 770 Nm−1

• Compression in the spring:d = 0.16 m

a.

Let v be the speed of the baseball when it is released i.e. when it is at its equilibrium position. As there are no dissipative forces involved, total mechanical energy of the system remains conserved. Therefore,

1/2 k×d^2 = m×g×d + 1/2 m×v^2

770×0.16^2 = 0.16*2×9.81×0.14 + 0.14×v^2⟹

V=11.733m/s

b.

Let h be the maximum height, above the equilibrium position, the baseball reaches. From the conservation of mechanical energy, this corresponds to the position where the initial elastic potential energy is completely converted into gravitational potential energy. Therefore,

m×g×(d + h) = 1/2 k×d^

0.14×9.81×(0.16+ h) = 0.5×770×0.16^2

h=7.016m

c.

Let v′ be the speed of the ball when it is at a height of h′ = 0.5×h = 0.5×7.016 = 3.508m above the equilibrium point. From the conservation of mechanical energy,

1/2 m×v′^2 + m×g×(d + h′) = 1/2 k×d^2

So v'=8.296m/s