Spring Constant Homework: Find Block's Speed, Compression, and Height

[Jtappan]

Homework Statement

A 2.9 kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spring is attached to a wall, as shown. The initial height of the block is 0.32 m above the lowest part of the slide and the spring constant is 442 N/m.

(a) What is the block's speed when it is at a height of 0.25 m above the base of the slide?
________ m/s
(b) How far is the spring compressed?
_______ m
(c) The spring sends the block back to the left. How high does the block rise?
_________ m

Homework Equations

PEi+KEi=PEf+KEf

and

PEi+KEi+1/2Kxi^2=PEf+KEf+1/2Kxf^2

The Attempt at a Solution

For A,

How do you set up the equations so you find the final velocity at that point?

 

[G01]

For A,

How do you set up the equations so you find the final velocity at that point?

You have pointed out that the conservation of energy equations are important for this problem, and you are correct. Now, can you tell me what the relationship is between the mechanical energy at .32m and the mechanical energy at .35m? If you can, you should have the basics for an equation that can help you find the velocity at .25m.

 

[ogb p]

To find the block's speed when it is at a height of 0.25 m above the base of the slide, we can use the conservation of energy equation: PEi + KEi = PEf + KEf.
Since the block is released from rest, its initial kinetic energy is 0. The initial potential energy is equal to the mass of the block multiplied by the acceleration due to gravity (g=9.8 m/s^2) and the initial height (0.32 m). So, PEi = mgh = (2.9 kg)(9.8 m/s^2)(0.32 m) = 9.05 J.
At a height of 0.25 m, the block has a potential energy of PEf = mgh = (2.9 kg)(9.8 m/s^2)(0.25 m) = 7.15 J.
Using the conservation of energy equation, we can solve for the final kinetic energy: KEf = PEi - PEf = 9.05 J - 7.15 J = 1.9 J.
Since KE = 1/2mv^2, we can solve for the final velocity: v = √(2KE/m) = √(2(1.9 J)/(2.9 kg)) = 1.27 m/s.
Therefore, the block's speed when it is at a height of 0.25 m above the base of the slide is 1.27 m/s.

For B,

To find how far the spring is compressed, we can use the equation: PEi + KEi + 1/2Kxi^2 = PEf + KEf + 1/2Kxf^2.
Since the block is released from rest, its initial kinetic energy is 0. The initial potential energy is equal to the mass of the block multiplied by the acceleration due to gravity (g=9.8 m/s^2) and the initial height (0.32 m). So, PEi = mgh = (2.9 kg)(9.8 m/s^2)(0.32 m) = 9.05 J.
At a height of 0.25 m, the block has a potential energy of PEf = mgh = (2.9 kg)(9.8 m/s^2)(0.25