How do I solve for total energy using kinetic and potential energy equations?

Click For Summary

Homework Help Overview

The discussion revolves around calculating total energy using kinetic and potential energy equations in the context of a problem involving a child sliding down a slope. Participants explore the relationship between energy conservation, friction, and the equations governing kinetic and potential energy.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the total energy equation and its components, questioning the impact of friction on energy calculations. Some explore the implications of using different values for gravitational acceleration and frictional force.

Discussion Status

The conversation includes various attempts to derive the velocity of the child at the bottom of the slope, with some participants providing algebraic manipulations and others questioning the assumptions made about friction. There is no explicit consensus on the final answer, but several productive lines of reasoning have been presented.

Contextual Notes

Participants note that the frictional force is defined as a fraction of the child's weight, which leads to discussions about the correct interpretation of the problem's parameters. The original poster's calculations and assumptions are also under scrutiny, particularly regarding the values used for gravitational acceleration and the effects of friction.

xRadio
Messages
60
Reaction score
0

Homework Equations



Total Energy = Kinetic Energy + Potential Energy
T.E = 1/2mv^2 + m(g)(h)

The Attempt at a Solution



Distance Child travels
Sin 30 = 4/ x
x = 8m

T.E at top
= 1/2mv^2 + m(9.8)(h)
= 40m
 
Last edited:
Physics news on Phys.org
Conservation of energy only works for non-conservative forces! You could do E = Work_net = Work_conservative + Work_non-conservative, but going back to Newton's 2nd law should give an easier time.
 
well you're on the right track, and assume from your last calculation that you're rounding g to 10m/s^2.

TE at bottom=pe+ke where pe=0

however, there is energy lost in going from top to bottom in form of friction.

so TE at top=TE at bottom plus frictional energy. You are given a magnitude for friction and have computed the distance right, can you finish from here?

edit: this was more or less simultaneous post, I think its actually easier using energy eqn, but solveable from either approach.
 
40m = 1/2mv^2 + 1/4m
v = 12.6 - m

is this correct?

And I don't know how to calculate the mass.
 
Last edited:
closer. this is how I approached the problem, and always BTW much better to complete the algebra before posting numbers--both for purposes here and on an exam as a simple number mistake made early will cost dearly;

mgh=1/2mv^2+Ff*distance where Ff=frictional force

we know from problem, that frictional force = 1/4mg

so mgh-1/4*(mg)*8m= 1/2mv^2
(8m from your calculations involving length of slide)
 
This is what I got:
40m-1/4*(m)*8m= 1/2mv^2
40m - 2m = 1/2mv^2
76m / m = v^2
v = 8.7 m/s

Question stated that friction is 1/4m not 1/4mg. Or is it suppose to be 1/4mg?
 
The question said the frictional force was one quarter of the child's weight. Weight is m*g, expressed in Newtons. So it should be (1/4)mg.
 
This is what I got now:
40m-1/4*(m)*8m= 1/2mv^2
40m - 20m = 1/2mv^2
20m / m = v^2
v = 4.47 m/s
 
Where did the 1/2 from the kinetic energy go? You dropped it in the second last line.
 
  • #10
O you thanks I forget to multiply by 2.

This is what I got now:
40m-1/4*(m*g)*8m= 1/2mv^2
40m - 20m = 1/2mv^2
40m / m = v^2
v = 6.3 m/s
 
  • #11
I agree with that answer.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
1K
Replies
11
Views
1K
Replies
6
Views
1K
Replies
55
Views
6K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K