Question

A spring has a force constant of 40,000 n/m. It is resting horizontally on a 20 m tall roof top and fires a 2 kg mass to the ground below. The mass hits the ground a 21 m/sec. How far was the spring compressed? (no energy was lost)

In the previous problem, if the mass hits the ground at 19 m/sec, how much energy was lost due to air resistance?

Answer 1

1) 4.9 cm

The initial mechanical energy of the spring+mass system is sum of gravitational potential energy and elastic potential energy of the spring:

$E=mgh+\frac{1}{2}kx^2$

where:

m = 2 kg is the mass

g = 9.8 m/s^2

h = 20 m is the height of the roof top

k = 40,000 N/m is the spring constant

x is the compression of the spring

When the mass hits the ground, its mechanical energy is just kinetic energy:

$E=\frac{1}{2}mv^2$

where v = 21 m/s is the speed. Since energy is conserved, we can equalize the two expressions, and solving for x we find the compression of the spring:

$mgh+\frac{1}{2}kx^2=\frac{1}{2}mv^2\\x = \sqrt{\frac{mv^2-2mgh}{k}}=\sqrt{\frac{(2 kg)(21 m/s)^2-2(2 kg)(9.8 m/s^2)(20 m)}{40000 N/m}}=0.049 m = 4.9 cm$

2) 79 J

The initial mechanical energy of the spring-mass system is

$E_i=mgh+\frac{1}{2}kx^2=(2 kg)(9.8 m/s^2)(20 m)+\frac{1}{2}(40,000 N/m)(0.049 m)^2=440 J$

While the final mechanical energy of the mass is

$E_f = \frac{1}{2}mv^2 = \frac{1}{2}(2 kg)(19 m/s)^2=361 J$

So, the energy lost due to air resistance is

$\Delta E= E_i - E_f = 440 J-361 J=79 J$