Block compresses spring, find coefficient of kinetic friction

[Nick-]

You push a 2.5 kg block against a horizontal spring, compressing the spring by 18 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 90 cm from where you released it. The spring constant is 200 N/m. What is the block-table coefficient of kinetic friction?



Having trouble finding the initial and final mech energy (sum of kinetic and elastic potential energy - I don't know how to find the velocity so I can't find the kinetic energy, would you set up energy equations?)



Using the spring constant I found: F=-k(d) = -200N/m (.18m) = 36 N
Elastic potential energy = 1/2kx^2 = 1/2(450N/m)(.18m)^2 = 7.29 N
The change in thermal energy = Fk(d) = (Uk)(Fn)(d)
Normal Force = mg = 24.5 N
Also, initial mech energy - change in thermal energy = final mech energy
That's pretty much all I got, any help would be much appreciated, thanks

 

[LowlyPion]

You push a 2.5 kg block against a horizontal spring, compressing the spring by 18 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 90 cm from where you released it. The spring constant is 200 N/m. What is the block-table coefficient of kinetic friction?

Having trouble finding the initial and final mech energy (sum of kinetic and elastic potential energy - I don't know how to find the velocity so I can't find the kinetic energy, would you set up energy equations?)

Using the spring constant I found: F=-k(d) = -200N/m (.18m) = 36 N
Elastic potential energy = 1/2kx^2 = 1/2(450N/m)(.18m)^2 = 7.29 N
The change in thermal energy = Fk(d) = (Uk)(Fn)(d)
Normal Force = mg = 24.5 N
Also, initial mech energy - change in thermal energy = final mech energy
That's pretty much all I got, any help would be much appreciated, thanks

Welcome to PF.

You know how much potential energy is in the spring just before it's released.

So what force acting over the 90 cm distance will have absorbed the PE that went into the object's KE? Won't that work equal the Potential it had when it was on the spring?

 

[thomate1]

.



Based on the given information, we can use the principle of conservation of energy to find the coefficient of kinetic friction. The initial mechanical energy of the block-spring system is equal to the final mechanical energy of the block-table system. This can be expressed as:

Initial mechanical energy = Final mechanical energy

The initial mechanical energy includes both the elastic potential energy stored in the compressed spring and the kinetic energy of the block as it is released from the spring. The final mechanical energy includes only the kinetic energy of the block as it slides across the table. We can write these equations as:

Initial mechanical energy = Elastic potential energy + Kinetic energy (1)

Final mechanical energy = Kinetic energy (2)

We can use the given information to solve for the elastic potential energy and the kinetic energy in these equations. The elastic potential energy can be calculated using the formula 1/2kx^2, where k is the spring constant and x is the compression distance. Plugging in the values, we get:

Elastic potential energy = 1/2(200N/m)(0.18m)^2 = 3.24 J

To find the kinetic energy, we can use the formula 1/2mv^2, where m is the mass of the block and v is the velocity of the block. We can use the fact that the block stops after sliding 90 cm to calculate its velocity. We know that the distance traveled is equal to the average velocity multiplied by the time, and the time can be calculated using the formula d = (1/2)at^2, where a is the acceleration. In this case, the acceleration is due to the force of friction, which is equal to the normal force multiplied by the coefficient of kinetic friction. Putting these equations together, we get:

90 cm = (1/2)v(t) (3)

and

a = μkFn (4)

Substituting (4) into (3) and solving for the time, we get:

t = √(2d/a) = √(2(0.9m)/(μkFn)) = √(1.8/μk(24.5N)) = 0.06/μk (5)

Now, we can use (5) to solve for the kinetic energy in equation (2):

Final mechanical energy = Kinetic energy = 1/2mv^2 = 1/2(2