How much work is needed to push a 100. -kg packing crate a distance of 10.0 m up a frictionless inclined plane that makes an angle of 30. degrees with the horizontal?
a. 4900 J
b. 2450 J
c. 1170 J
d. 2210 J
2006-07-31 07:38:42 · 5 answers · asked by todd b 1 in Science & Mathematics ➔ Mathematics
Well, the force is gravity with g=9.8 m/s^2. If you go 10m along an incline of 30 degrees, you go up a distance of 10*sin(30degrees)=10*.5=5m. Now do mgh: m=100kg, g=9.8 m/s^2, and h=5m, so
mgh=100*9.8*5=4900 J.
2006-07-31 07:45:55 · answer #1 · answered by mathematician 7 · 1⤊ 0⤋
First we have to figure out where the work is being done. The key here is that the plane is frictionless. That means that there will be no work done horizontally, nor along the path of the incline. Thus, work is only done in the vertical direction, against gravity.
To figure out the work done, we simply multiply force and distance. Here, the force is going to be the weight of the object, m*g. Distance is simply the hight from the bottom of the incline to tohe top of the incline.
Thus, height is going to be
10 sin (30) = 5 m
thus
W = m*g*h
m = 100, g = 9.81
W = 100*9.81*5
= 4905 Joules
2006-07-31 08:09:43 · answer #2 · answered by polloloco.rb67 4 · 0⤊ 0⤋
Others have done this, but I'll do it too. Work is force x distance. The horizontal distance doesn't matter (no force needed), and the vertical distance is 10 sin 30 = 5 m.
F = ma = -mg = -(100 kg)(-9.8 m/s/s) = 980
W = 980 x 5 = 4900 J.
2006-07-31 09:36:12 · answer #3 · answered by bpiguy 7 · 0⤊ 0⤋
If it is on the surface of earth, it is around 4900 J.
m = 100 kg
g = 9.8 m/s2
h = 10 * sin 30deg = 5 m
work = m*g*h = 4900 J
2006-07-31 09:24:45 · answer #4 · answered by Anonymous · 0⤊ 0⤋
D
2006-07-31 07:42:17 · answer #5 · answered by sparkloom 3 · 0⤊ 0⤋