F.3 Phy 1條問題

is it 60 cm?

As the box is lifted at constant velocity, the word done by the man

= G.P.E. gained by the box

= mgh

= (5)(10)(0.6)

= 30 J


b. Work done on the box = -mgh = -30 J

這裏可以幫到你
http://actioni.w7.c361.net/yahoo.com.hk/hk/auction/178987536

1. First choice:
By Newton's first law of motion,
Force pulling the box upward = weight of the box (constant velocity)
By W = mg,
weight = 5 x 10 = 50N
By W = Fs,
work done = 50 x 0.6 = 30J

Second choice:
Work done = PE gained by the box
By PE = mgh,
Work done = (5)(10)(0.6)
= 30J

2. a)
First choice:
By W = mg,
weight of the box = 5 x 10 = 50N
By W = Fs,
work done = 50 x 0.6 = 30J

Second choice:
Work done = loss of PE
By PE = mgh,
work done = 5 x 10 x 0.6 = 30J

Third choice:
u = 0
v?
a = 10 ms^-2
s = 0.6

By v^2 - u^2 = 2as,
v = sqrt 12

Work done = gain of KE,
By KE = 0.5mv^2,
Work done = 0.5(5)(sqrt 12)^2 = 30J

1.As the box is lifted at a constant velocity .
thus force appling on the box=weight of the box
=5(10) (As W=mg)
=50N


the work done by the man during the lifting=(force appling on the box)(displacement
travelled by the box)

=50(60/100)

=30J


2.(a) when the box fall steadily back to the ground,the force act on the box by the

man=weight of the box in magnitude,but direction of the force act on the box by the

man is oposite to that of the displacement of the box

the work done on the box by the man =(force appling on the box)(displacement
travelled by the box)


=-50(60/100)
=-30J


the work done on the box by the weight=(weight)(displacement
travelled by the box)


=50(60/100)
=30J ( direction of the weight is the same
as the displacement
travelled by the box)