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A 20 kg box is attached to a compressed spring that has a spring constant of 300 N/m. the box is resting on a frictionless surface and the spring is compressed 30 cm.

A. what is the EPE of spring?


B. what will be the KE of the box when the spring expands back to its natural length?

C. how fast will the box be moving after the spring releases the box?

User Dain
by
3.1k points

2 Answers

23 votes
23 votes

Final answer:

The spring's Elastic Potential Energy (EPE) is 13.5 J; this is fully converted into the box's Kinetic Energy (KE) when the spring expands. The box moves at approximately 1.16 m/s after the spring releases it.

Step-by-step explanation:

Understanding Elastic and Kinetic Energy


Let's go through each part of the problem step-by-step.

A. Elastic Potential Energy (EPE) of the Spring


The Elastic Potential Energy (EPE) stored in a compressed spring can be calculated using the formula EPE = (1/2)kx^2, where k is the spring constant and x is the compression distance. In this case, k = 300 N/m and x = 30 cm (0.30 m). Plugging in the values, we get EPE = (1/2) * 300 N/m * (0.30 m)^2 = 13.5 J (Joules).

B. Kinetic Energy (KE) of the Box


When the spring returns to its natural length, all of the EPE is converted into the box's Kinetic Energy (KE). Thus, the KE of the box will also be 13.5 J, as energy is conserved in this frictionless system.

C. Velocity of the Box


To find the velocity of the box after the spring releases it, we use the formula KE = (1/2)mv^2, where m is the mass of the box and v is the velocity. We can solve for v by rearranging the formula: v = sqrt((2 * KE) / m). With m = 20 kg and KE = 13.5 J, the velocity v = sqrt((2 * 13.5 J) / 20 kg) = sqrt((27 J) / 20 kg) = sqrt(1.35) m/s ≈ 1.16 m/s. So, the box will be moving at approximately 1.16 m/s after the spring releases it.

User Bbengfort
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2.5k points
14 votes
14 votes

Answer:

Below

Step-by-step explanation:

A) Elastic Potential Energy = 1/2 k (d^2) = 1/2 * 300 N/m * .3^2 m =13.5 N m

B) The EPE will be ransferred to KE of the box = 13.5 J

C) KE = 13.5 = 1/2 m v^2

= 1/2 (20)(v^2 ) shows v = 1.16 m/s

User Adam Lenda
by
2.7k points