Finding the Transferred Energy for a Block Sliding Down a Ramp

[greenglasses]

Homework Statement

A 5.0kg block slides down a ramp, starting with a velocity down the slope of 2.5 m/s. The ramp is 1.5 m high and has an angle of 25 degrees. The force of friction acting upon the block is 20.0 N.

How much energy is transferred in or out by gravity, normal force, and friction respectively as it slides down the ramp?

Homework Equations

Ek = 1/2 mv²
Ep = mgh
Fg = 9.8m
Fn = Fg_x
Ff =μFn

The Attempt at a Solution

This question is actually a five part question. I have completed parts 1 and 5, which is here in case it may be useful:
1: How much energy does the block begin with?
A: Ek_i= 0.5(5.0kg)(2.5²) = 15.6 J
5. What is the final velocity of the block?
Ef = Ei + W_Ff
(0.5)(5.0kg)v_f² = (5.0kg)(9.8m/s/s)(1.5m) + (0.5)(5.0kg)(2.5²m/s) - (20N)(1.5/cos(25)m)
v_f = 2.69 m/s

I'm not sure what concept I'm supposed to apply to discover "energy transferred", however. I thought that the energy within a closed system remained constant. I do not want the answers; I simply want some sort of push in the right direction.

 

[Dick]

Homework Statement

A 5.0kg block slides down a ramp, starting with a velocity down the slope of 2.5 m/s. The ramp is 1.5 m high and has an angle of 25 degrees. The force of friction acting upon the block is 20.0 N.

How much energy is transferred in or out by gravity, normal force, and friction respectively as it slides down the ramp?

Homework Equations

Ek = 1/2 mv²
Ep = mgh
Fg = 9.8m
Fn = Fg_x
Ff =μFn

The Attempt at a Solution

This question is actually a five part question. I have completed parts 1 and 5, which is here in case it may be useful:
1: How much energy does the block begin with?
A: Ek_i= 0.5(5.0kg)(2.5²) = 15.6 J
5. What is the final velocity of the block?
Ef = Ei + W_Ff
(0.5)(5.0kg)v_f² = (5.0kg)(9.8m/s/s)(1.5m) + (0.5)(5.0kg)(2.5²m/s) - (20N)(1.5/cos(25)m)
v_f = 2.69 m/s

I'm not sure what concept I'm supposed to apply to discover "energy transferred", however. I thought that the energy within a closed system remained constant. I do not want the answers; I simply want some sort of push in the right direction.

I think the "system" they want you to think about is just the block, and that system is not closed. You know the initial and final kinetic energies of the system. They are asking about the net effect of each of those three forces to the kinetic energy of the block. Since you are going down, gravity can only put energy in. Friction can only take energy out. Can normal force do either? The answers to those are in the way you computed vf. Is that an ok push?

 

[greenglasses]

Oh, I think I get it now. So the energy gravity put in would just be the amount of energy converted from potential to kinetic, and the energy transferred for friction would just be Ff*Δs.

I can't think of the effect of normal force... I'm guessing you're implying that the answer for that part is zero joules, then?

Thank you for your help. (I'd appreciate it if you informed me if I misinterpreted anything, however.)

 

[Dick]

Oh, I think I get it now. So the energy gravity put in would just be the amount of energy converted from potential to kinetic, and the energy transferred for friction would just be Ff*Δs.

I can't think of the effect of normal force... I'm guessing you're implying that the answer for that part is zero joules, then?

Thank you for your help. (I'd appreciate it if you informed me if I misinterpreted anything, however.)

You are interpreting everything just fine. As for the normal force, it's called "normal" because it's always "normal" = "perpendicular" to the direction of motion. Can such a force ever transfer energy to or from an object?

 

[HalfLostAndSearching]

In order to find the energy transferred by each force, we can use the conservation of energy principle, which states that the total energy in a closed system remains constant. However, energy can be transferred from one form to another within the system. In this case, the system includes the block, the ramp, and the forces acting upon the block.

To find the energy transferred by gravity, we can use the equation Ep = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the ramp. Using the given values, we can calculate the potential energy at the top of the ramp to be 73.5 J. At the bottom of the ramp, the potential energy is 0 J, so the difference between these two values (73.5 J - 0 J = 73.5 J) represents the energy transferred by gravity.

To find the energy transferred by the normal force, we can use the equation Fn = Fgx, where Fg is the force of gravity and x is the distance along the ramp. Since the normal force and the force of gravity are equal in magnitude but opposite in direction, the energy transferred by the normal force is the same as the energy transferred by gravity.

To find the energy transferred by friction, we can use the equation Ff = μFn, where μ is the coefficient of friction and Fn is the normal force. Again, since the normal force and the force of gravity are equal in magnitude but opposite in direction, the energy transferred by friction is the same as the energy transferred by gravity.

Overall, the total energy transferred by all forces is 73.5 J + 73.5 J + 73.5 J = 220.5 J. This represents the change in kinetic energy of the block as it moves from the top to the bottom of the ramp. We can use this value to check our work by calculating the final kinetic energy using the equation Ek = 1/2 mv2, where m is the mass of the block and v is the final velocity. Plugging in the given values, we get 0.5(5.0 kg)(2.692 m/s) = 20.2 J, which is close to the expected value of 15.6 J calculated in the first part of the problem.

In summary, to find the energy transferred by each force, we can use the appropriate equations and consider the