Finding the Transferred Energy for a Block Sliding Down a Ramp

Click For Summary

Homework Help Overview

The problem involves a 5.0 kg block sliding down a ramp with an initial velocity of 2.5 m/s, a height of 1.5 m, and an angle of 25 degrees. The block experiences a frictional force of 20.0 N. The question seeks to determine the energy transferred by gravity, normal force, and friction as the block descends the ramp.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial and final kinetic energies, the role of gravitational potential energy, and the effects of friction. There is an exploration of how to quantify the energy transferred by each force acting on the block.

Discussion Status

Some participants have provided insights into the energy contributions from gravity and friction, while questioning the role of the normal force in energy transfer. There is an acknowledgment of the need for clarification on the concept of energy transfer within the context of the problem.

Contextual Notes

Participants note that the system is not closed due to the presence of external forces, and there is uncertainty about how to account for the normal force in terms of energy transfer.

greenglasses
Messages
23
Reaction score
0

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 = Fgx
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: Eki= 0.5(5.0kg)(2.52) = 15.6 J
5. What is the final velocity of the block?
Ef = Ei + WFf
(0.5)(5.0kg)vf2 = (5.0kg)(9.8m/s/s)(1.5m) + (0.5)(5.0kg)(2.52m/s) - (20N)(1.5/cos(25)m)
vf = 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.
 
Physics news on Phys.org
greenglasses said:

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 = Fgx
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: Eki= 0.5(5.0kg)(2.52) = 15.6 J
5. What is the final velocity of the block?
Ef = Ei + WFf
(0.5)(5.0kg)vf2 = (5.0kg)(9.8m/s/s)(1.5m) + (0.5)(5.0kg)(2.52m/s) - (20N)(1.5/cos(25)m)
vf = 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?
 
Last edited:
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.)
 
greenglasses said:
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?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 45 ·
2
Replies
45
Views
8K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
14
Views
3K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 4 ·
Replies
4
Views
8K
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
7K
Replies
18
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K