gravitational potential energy

Crystal

2012-02-26 9:59 pm

圖片參考：http://imgcld.yimg.com/8/n/HA04937267/o/701202260040213873441670.jpg

sor.....幅圖可能有D唔清,因為要遮住d notes...
上面幅圖係解釋點解PE=mgh

1.佢係assume左lifting force = weight (mg)
咁如果lifting force 唔等於 mg 呢?(eg 大過mgh)
點解PE都係=mgh?

2.lifting force balanced 左個weight.
Take upward direction as positive
work done by weight on the block = Fs
= m(-g)h
=-mgh
如果係咁,即係無energy gain,咁點解PE會increases?

另外,仲有2題唔係好明......都係關於mgh
1.Peggy finds out her power by running up the stairs.Which of the following is
not required in calculating her power?
A.The time for her to run up the stairs
B.Her weight
C.The vertical height of the stairs
D.Total length of the stairs

我知D係錯,但我唔明點解B係岩
Take the starting point of Peggy as the reference level,where PE=0
gain in PE = work done by vertical component of forward force(instead of the weight)

By W=Fs ,F= vertical component of forward force,s= vertical displacement
W=gain in PE,
By P=W/t,差個time就可以搵到power
咁知佢個weight黎做咩?

更新1:

thx......... 第2題,The work done by the lifting force = (+mg).h =+mgh 咁咪岩岩好compensate左個 work done by weight = -mgh 咁點解佢會有gravitiational pe gain?

更新2:

typo....gravitational

更新3:

=.=......

回答 (1)

天同

2012-02-26 11:24 pm

✔ 最佳答案

1. When the lift force equals to the weight, we have assumed that the mass is being lifted with negligible velocity, i.e. very very slowly. All the work done by the force is transferred to the potential energy (PE) of the mass.

If the lifting force is larger than the weight, the weight would have acceleration and hence gain kinetic energy (PE). In this case, the work done by the lifting force = gain in PE + gain in KE. That is, work done by lifting force > gain in PE.

2. Weight of mass = -mg (i.e. it is pointing downward)
The lifting force must be acting upward in order to balance the weight, thus lifting force = +mg
The work done by the lifting force = (+mg).h =+mgh
Be aware that it is the work done by the lifting force, not the work done by the weight.
------------------------------------------
Just use your answer and I give my amendment in red :

Take the starting point of Peggy as the reference level,where PE=0
gain in PE = work done by vertical component of forward force(instead of the weight)

By W=Fs ,F= vertical component of forward force,
hence, F = mg

s= vertical displacement
W=gain in PE=( mg)s ,
By P=W/t,差個time就可以搵到power
咁知佢個weight黎做咩 Because weight = mg

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