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# gr 12 physics?

A 31.0 kg child on a swing reaches a maximum height of 1.92 m above their rest position. Assuming no loss of energy:

a) At what point during the swing will she attain their maximum speed?

b) What will be her maximum speed through the subsequent swing?

c) Assuming this maximum height was the result of one push from her parent, what was the average force exerted by the parent if they pushed over a distance of 152 cm?

### 4 Answers

- Andrew SmithLv 71 month agoFavorite Answer
When the potential energy is a minimum the kinetic energy is a maximum. ie at the bottom of the swing.

b) Ek = Ep -> 1/2 m v^2 = mgh-> v= sqrt( 2h*g) = sqrt( 2*1.92*9.8) m/s

c) E = F.d -> mgh = F.d -> F = mgh/d= 31*9.8*1.92/1.52 N

- Jim MoorLv 71 month ago
Conservation of Energy

When swing is at the lowest, the PE has become KE

mgh = 1/2 m v^2

So v = √(2gh) so plug in numbers and solve, round to 3 sigfigs

6.14 m/s

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- Anonymous1 month ago
a)

at the lowest point of the swing

b)

V = √2gh = √19.612*1.92 = 6.14 m/sec

c)

m*g*h = F*d

F = 31*9.806*1.92/1.52 = 384 N

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thanks .. put in a division when it should have been a multiplication. fixed now.