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

AI Thread Summary
To solve for total energy using kinetic and potential energy equations, the relationship is established as Total Energy (T.E) equals Kinetic Energy (KE) plus Potential Energy (PE). The discussion emphasizes the importance of accounting for energy lost due to friction when transitioning from top to bottom of a slide. The calculations involve setting up the equation mgh - frictional force = 1/2mv^2, with friction defined as one-quarter of the child's weight. After several iterations and corrections, the final velocity is calculated to be approximately 6.3 m/s. The conversation highlights the necessity of careful algebraic manipulation to avoid errors in physics problems.
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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
 
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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.
 
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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.
 
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