How Much Elastic Potential Energy Was Stored in the Spring Before Release?

[kevykevy]

Question-A child placed a spring of negligible mass between two toy cars of masses 0.112 kg and 0.154 kg. She compresses the spring and ties the cars together with a piece of string. When she cuts the string, the spring is released and the cars move in opposite directions. The 0.112 kg car has a speed of 1.38 m/s. What was the elastic potential energy stored in the spring before the string was cut?

Solution-I've tried to start this and I have no idea even where to begin. All I have so far is this picture I drew describing what's happening...

Spring Compressed
----

Spring Uncompressed (at rest)
- - - -

Before
Car1----Car2

After
Car1- - - -Car2

 

[OlderDan]

Question-A child placed a spring of negligible mass between two toy cars of masses 0.112 kg and 0.154 kg. She compresses the spring and ties the cars together with a piece of string. When she cuts the string, the spring is released and the cars move in opposite directions. The 0.112 kg car has a speed of 1.38 m/s. What was the elastic potential energy stored in the spring before the string was cut?

Solution-I've tried to start this and I have no idea even where to begin. All I have so far is this picture I drew describing what's happening...

Spring Compressed
----

Spring Uncompressed (at rest)
- - - -

Before
Car1----Car2

After
Car1- - - -Car2

What physical quantities do you think are conserved in this process?

 

[kyle8921]

To solve this problem, we will use the equation for elastic potential energy:

Elastic potential energy = 1/2 * k * x^2

Where k is the spring constant and x is the displacement of the spring from its original position. We can calculate the spring constant using the equation:

k = F/x

Where F is the force applied on the spring and x is the displacement.

In this case, the force applied on the spring is equal to the force applied on both cars, which is equal to their combined mass (0.112 kg + 0.154 kg) multiplied by their acceleration (1.38 m/s^2). This gives us a force of 0.359 N.

Now, we need to find the displacement of the spring. From the given information, we know that the 0.112 kg car has a speed of 1.38 m/s after the string is cut. Using the equation for kinetic energy (KE = 1/2 * m * v^2), we can calculate the kinetic energy of the car, which is also equal to the elastic potential energy stored in the spring.

KE = 1/2 * 0.112 kg * (1.38 m/s)^2 = 0.107 J

We also know that the kinetic energy of the car is equal to the elastic potential energy stored in the spring, so we can set up an equation:

Elastic potential energy = 0.107 J = 1/2 * k * x^2

We already know the force applied on the spring (0.359 N), so we can rearrange the equation to solve for x:

x = √(0.107 J * 2 / 0.359 N) = 0.293 m

Now, we can plug in the values for k and x into the original equation for elastic potential energy to find the answer:

Elastic potential energy = 1/2 * 0.359 N * (0.293 m)^2 = 0.0156 J

Therefore, the elastic potential energy stored in the spring before the string was cut is 0.0156 J.