Power and energy

用戶225241

2008-01-16 12:24 am

A man pushes a 45kg block along a plank from the ground to a lorry as shown above. The frictional force between the block and the plank is 30N. The block moves with a uniform velocity of 40cm/s.
the plank is 3m oblique , 1.2cm 橫
a) Find the force exerted by the man
b) What is the work done by the man
c) Find the work done against friction
d) What is the potential energy gained by the block?
e) What is the average power developed by the man?

回答 (1)

Audrey Hepburn

2008-01-16 6:26 am

✔ 最佳答案

a. Angle of the plank = cos-1(1.2 / 3) = 66.422˚

Force exerted by the man, F

= Frictional force + parallel component of the weight of the block

= f + mgsinθ

= 30 + 450sin66.422˚

= 442.4 N

b. Work done by the man

= Force exerted by the man X displacement

= 442.4 X 3

= 1327 J

c. Work done against friction

= Frictional force X displacement

= 30 X 3

= 90 J

d. Gain in height = [(3)2 – (1.2)2]1/2 = 2.75 m

Potential energy gained by the block

= mgh

= 450(2.75)

= 1237 J

e. Average power

= Force X velocity

= 442.4 X 0.4

= 177 W

f. Gain in K.E. + work done against friction = Lost in P.E.

1/2 mv2 + 90 = 1237

1/2 (45)v2 = 1147

v = 7.14 ms-1

The speed of the block at the bottom of the plank is 7.14 ms-1.

參考: Myself~~~

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