In: Physics
A factory worker pushes a 32.0 kg crate a distance of 4.3 m along a level floor at constant velocity by pushing downward at an angle of 29 ? below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.
part a. What magnitude of force must the worker apply to move the crate at constant velocity? (N)
part b. How much work is done on the crate by this force when the crate is pushed a distance of 4.3 m ? (J)
part c. How much work is done on the crate by friction during this displacement? (J)
part d. How much work is done by the normal force? (J)
part e. How much work is done by gravity? (J)
part f. What is the total work done on the crate? (J)
(a) What magnitude of force must the worker apply to move the crate at constant velocity?
fk = k (m g + F sin )
F cos = k (m g + F sin )
F cos 290 = (0.25) [(32 kg) (9.8 m/s2) + F sin 290]
(0.8746) F = (78.4 N) + (0.1212) F
F = (78.4 N) / [(0.8746) - (0.1212)]
F = [(78.4 N) / (0.7534)]
F = 104.06 N
(b) How much work is done on the crate by this force when the crate is pushed a distance?
we know that, W = F d cos
W = (104.06 N) (4.3 m) cos 290
W = 391.3 J
(c) How much work is done on the crate by friction during this displacement?
Since there is no net force acting on the crate, then the net work done on crate is 0.
Therefore, we have
Wfriction = - 391.3 J
(d) How much work is done by the normal force?
The normal force acts in a direction that is perpendicular to the displacement vector.
Then, we get
Wnormal = 0 J
(e) How much work is done by gravity?
Gravity acts in a direction that is perpendicular to the displacement vector.
Then, we get
Wgravity = 0 J
(f) What is the total work done on the crate?
Wtotal = Wforce + Wfriction + Wnormal + Wgravity
Wtotal = [(391.3 J) + (-391.3 J) + (0 J) + (0 J)]
Wtotal = 0 J