Elastic Potential Energy Problem

Question

A linear spring (k=8.5 kN/m) is compressed 22 cm before being released to launch a 55-g steel ball upward at an angle of 30 degrees from horizontal. Ignoring friction and other losses, and assuming g=9.81 m/s^2, what is the time needed for the ball to reach maximum height?

Answer 1

potential energy (PE) is force times distance. In the case of a spring, the force is proportional to the distance: F =k*d, where k is the spring constant. So the PE in a compressed spring is k*d^2

You have k = 8,500 Newtons/meter and d = 22 cm (0.22 m)

So PE = 8,500 * 0.22^2 = 411.4 Joules

When the ball is launched, the PE is converted to Kinetic Energy (KE), which is 1/2 * m * v^2

We are given m = 55 gm = 0.055 kg, so we can write the equation:

KE = PE = 411.4 = 1/2 * 0.055 * v^2

Rearrange this and solve for v.

Next, find the vertical component of v: v(v) = v * sin(30)

Once you have v(v), then use v = a*t to find the time to reach max. height (where the velocity drops to 0)

Answer 2

At time t₀, we can draw the situation like this:

Where the total energy (E_T) is just the spring potential energy 1/2kx² = 1/2(8.5x10³)(0.22)² = 205.7 joules

At time t₁, after the spring has been released and the ball is at the end of the spring travel (so that the spring is 100% uncompressed) we can draw the situation like this:

where the total energy is the kinetic energy plus the small amount of potential energy due to the x₁ height. So E_T = 1/2mv² + mgh where h = x₁ = 0.22sin(30)

205.7 = 1/2(0.055)v² + (0.055)(9.81)(0.22sin(30)

v = 86.475 m/s

Then v_y = 86.475sin(30) = 43.237 m/s (we could have ignored the small potential energy because it doesn't affect the results within the sig figs we are using)

To find the time to the max height, (where v_y = 0), use the kinematic equation v_f = v_i + at. Solving for t,:

t = (v_f - v_i)/a = (0 - 43.237)/(-9.81) = 4.4 sec