A worker pushes a 1.50x10^3 N crate with a horizontal force of 345 N a distance of 24.0m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220.
b) How much work is done by the floor on the crate?
Uk=Fk/Fn Fk=0.220; Fk=? ; Fn= 1.50x10^3 N
Fk=330N
W=F(d)
W=(-15N)(24.0m)
W= -3.6x10^2 J
Right?
2006-12-10 16:46:52 · 4 answers · asked by billf39 2 in Science & Mathematics ➔ Physics
The work done by the floor on the crate is the work done by the frictional force alone.
The frictional force is the product of the coefficient of kinetic friction and normal reaction of the floor. Normal reaction = weight of the body.
Therefore, the frictional force is 0.22 x 1.5 x 10^3 = 330 N.
Your answer for Fk is correct.
Work done by this friction is 330 x distance moved = 330 x 24 = 7920 J.
2006-12-10 17:52:52 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
The floor pushes against the crate with the same force as the friction force, as long as the crate is not accelerating. But the floor does not move. Work is done by the moving force. That would be 24m*330N
2006-12-10 17:13:45 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋
I would say the floor is absorbing 330x24 Nm of energy and the force is doing 345x24 Nm of work and the difference 15x24 Nm ends up as kinetic energy from which you can work out the velocity of the crate at 24m.
2006-12-10 17:01:46 · answer #3 · answered by nor^ron 3 · 1⤊ 0⤋
EDIT - I by no skill said the conveyor became vertical, my first reaction became incorrect. W=Fd like you've it, entire rigidity is the load plus the further friction, all accelerated by the gap travelled. (.0.5(9.eighty one)+.350)*a million.33=a million.44J You rounded off on your rigidity calc, it could be a million.088N no longer a million.0N, this gave you a million.33N particularly of the a million.44N.
2016-11-25 20:09:18 · answer #4 · answered by melgoza 4 · 0⤊ 0⤋