物理 ENERGY PE&KE

[匿名]

2008-10-25 10:41 pm

a ball of mass 50 g is dropped from rest at a height of 3m above the ground.
what is its speed just before reaching the ground?
if the speed of the ball just after the rebound is 4m/s . how much KE is lost?where does the energy go?
how high can the ball reach after the rebound?

回答 (1)

Audrey Hepburn

2008-10-25 10:47 pm

✔ 最佳答案

By the law of conservation of energy

Loss of G.P.E. = Gain of K.E.

mgh = 1/2 mv2

(10)(3) = 1/2 v2

Speed just before reaching the ground, v = 7.75 ms-1

K.E. just before hitting the ground

= 1/2 mv2

= 1/2 (0.050)(7.75)2

= 1.5 J

K.E. just after hitting the ground

= 1/2 mu2

= 1/2 (0.050)(4)2

= 0.4 J

Amount of K.E. lost = 1.5 - 0.4 = 1.1 J

The energy lost is dissipated as heat.

By the law of conservation of energy

Gain of G.P.E. = Loss of K.E.

mgh' = 0.4

(0.050)(10)h' = 0.4

Height after rebound, h' = 0.8 m

參考: Myself~~~

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