Question

The spring is compressed a total of 3.0 cm, and used to set a 500 gram cart into motion. Find the speed of the cart at the instant it is released, assuming all the elastic potential energy is converted.

Answer 1

Answer:

1.15 m/s

Explanation:

Part of the question is missing. Found the missing part on google:

"1. A hanging mass of 1500 grams compresses a spring 2.0 cm.   Find the spring constant in N/m."

Solution:

First of all, we need to find the spring constant. We can use Hooke's law:

$F=kx$

where

$F=mg=(1.5 kg)(9.8 m/s^2)=14.7 N$ is the force applied to the spring (the weight of the hanging mass)

x = 2.0 cm = 0.02 m is the compression of the spring

Solving for k, we find the spring constant:

$k=\frac{F}{x}=\frac{14.7}{0.02}=735 N/m$

In the second part of the problem, the spring is compressed by

x = 3.0 cm = 0.03 m

So the elastic potential energy of the spring is

$U=\frac{1}{2}kx^2=\frac{1}{2}(735)(0.03)^2=0.33 J$

This energy is entirely converted into kinetic energy of the cart, which is:

$U=K=\frac{1}{2}mv^2$

where

m = 500 g = 0.5 kg is the mass of the cart

v is its speed

Solving for v,

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.33)}{0.5}}=1.15 m/s$