A 0.075kg ball in a kinetic sculpture is raised 1.33 m above the ground by a motorized vertical conveyor belt. A constant frictional force of 0.350N acts in the direction opposite the conveyor belt's motion. How much total work is done in raising the ball?
What i did:
0.075 kg * 9.81m/s^2 = 0.73575 N = weight force of ball on earth
down forces= 0.736 N + 0.350N = 1N
Opposite force is 1N since acc is constant
Wnet = Fnet (d) (cos ??)
Wnet = (1N) (1.33m)
wnet = 1.33 J
RIGHT??
2006-12-10 16:16:20 · 4 answers · asked by billf39 2 in Science & Mathematics ➔ Physics
Whatever is the value of the frictional force and its direction, for a body to be lifted to a height h from the ground a work of mgh has to be done.
In your problem the lifting force is given only by the friction of the conveyor belt.
This gives a component force which is opposite to the weight of the ball.
In other words, the component of the friction gives the upward force to lift the ball which should be mg.
Hence the answer is 0.075 x 9.8 x 1.33 = 0.97755 Joule.
The conveyor belt acts like a simple machine only.
If we know the efficiency of the machine then we can find the actual work we did.
Or if the angle of frictional force to the horizontal is known then we can calculate the efficiency and or directly the total work done.
The frictional force cannot act vertically downward and if it acts so then the reaction force will act perpendicular to the weight of the ball.
2006-12-10 18:43:49 · answer #1 · answered by Pearlsawme 7 · 0⤊ 2⤋
EDIT - I never noticed the conveyor was vertical, my first response was incorrect.
W=Fd like you have it, total force is the weight plus the additional friction, all multiplied by the distance travelled.
(.075(9.81)+.350)*1.33=1.44J
You rounded off in your force calc, it should be 1.088N not 1.0N, this gave you 1.33N instead of the 1.44N.
2006-12-10 16:25:12 · answer #2 · answered by Mukluk 2 · 0⤊ 0⤋
Down force =0.075 kg*9.81 m/s^2 +0.35 N = 1.09 N
Distance traveled = 1.33 m
Work = force * distance = 1.09 N *1.33 m = 1.44 J
2006-12-10 16:44:25 · answer #3 · answered by arbiter007 6 · 0⤊ 0⤋
The flooring pushes in opposition to the crate with the identical drive because the friction drive, so long as the crate isn't accelerating. But the ground does now not transfer. Work is completed by means of the relocating drive. That could be 24m*330N
2016-09-03 08:37:17 · answer #4 · answered by ? 4 · 0⤊ 0⤋