How much work is needed to push a 132 kg (or 100.-kg) packing crate a distance of 5.00 m up a frictionless inclined plane that makes an angle of 20 degrees (30.0 degrees) with the horizontal?
a. 210 J
b. 2450 J
c. 1170 J
d. 2212 J
2006-08-10 08:37:57 · 6 answers · asked by todd b 1 in Science & Mathematics ➔ Physics
Since is's frictionless, it's an easy problem. The work you do translates directly into the increase in potential energy of the crate (this is called the 'principle of work energy equivalence')
5m*sin(20) = 1.71 m vertical component.
The potential energy is given by E=mgh (where m is mass, g is the gravitational accelleration of the Earth, and h is the distance above the zero reference)
Plug 'n play
W = 132*9.8*1.71 = 2,212 Joules of work
You do the other one
Doug
2006-08-10 08:51:36 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
Work done on the crate = force applied to the crate x distance the crate is pushed.
Since the crate is pushed up an incline, the force you apply needs to be just enough to overcome the force of gravity. ( It would be reduced by friction if the plane were not frictionless).
Since the plane IS inclined however, and you are not lifting the crate straight up over your head, gravity is only partially affecting the crate's motion:
F(gravity) = m*g = 9.8 [m/sec^2]* 132 [kg]
=1293.6 Newtons
That would be the force to counteract gravity by lifting the crate over your head, the force in the y-direction. At 20 degrees, the force to push the crate :
1293.6 * sin (20)
=442.44N
Work = F * d
=442.44 N * 5.00 m
=2212.18 J
or "D"
2006-08-10 18:45:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
i won't do your homework for you but i can point you in the right direction. use trig to find the height of the triangle, the hypotenuse is 5 m and the angle is 20 degrees (or 30? not sure what you meant. )
hint use sine.
work is determined by the amount an object is raised or lowered.
plug the numbers into your work equation and there ya go
2006-08-10 15:56:55 · answer #3 · answered by zaphods_left_head 3 · 1⤊ 0⤋
at 20 degrees, the inclined plane rises 2m if the hypotenus is 5m
cos(20) = A/H
H*cos(20) = A
5m * cos(20) = 2.04m
Work = force x length
Force to overcome gravity = 9.8 m/s on earth
1 KG = 9.8 Newtons weight on earth
132 KG = 1293.6 N weight
work = weight x distance
= 1293.6 x 2m = 2587 J ?
that doesn't match your choices...
2006-08-10 16:05:25 · answer #4 · answered by marwood0 2 · 0⤊ 0⤋
could it be 2212 joule
2006-08-10 16:34:25 · answer #5 · answered by problemsolver86 3 · 0⤊ 0⤋
A
2006-08-10 15:45:09 · answer #6 · answered by dbear 2 · 0⤊ 0⤋