- #1

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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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- #1

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Total Energy = Kinetic Energy + Potential Energy

T.E = 1/2mv^2 + m(g)(h)

Distance Child travels

Sin 30 = 4/ x

x = 8m

T.E at top

= 1/2mv^2 + m(9.8)(h)

= 40m

Last edited:

- #2

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- #3

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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.

- #4

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40m = 1/2mv^2 + 1/4m

v = 12.6 - m

is this correct?

And I dont know how to calculate the mass.

v = 12.6 - m

is this correct?

And I dont know how to calculate the mass.

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- #5

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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)

- #6

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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?

- #7

hage567

Homework Helper

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- #8

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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

40m-1/4*(m)*8m= 1/2mv^2

40m - 20m = 1/2mv^2

20m / m = v^2

v = 4.47 m/s

- #9

hage567

Homework Helper

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Where did the 1/2 from the kinetic energy go? You dropped it in the second last line.

- #10

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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

hage567

Homework Helper

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I agree with that answer.

- #12

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Do you think you can help me with my last problem?

https://www.physicsforums.com/showthread.php?t=164105

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