A factory worker pushes a 29.2kg crate a distance of 4.1m along a level floor at constant velocity by pushing downward at an angle of 28* below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.
(a) What magnitude of force must the worker apply?
(b) How much work is done on the crate by this force?
(c) How much work is done on the crate by frication?
(d) How much work is done on the crate by normal force?
(e) How much work is done on the crate by the gravity?
(f) What is the total work done on the crate?
A factory worker pushes a 30.4 kg crate a distance of 5.0 m along a level floor at constant velocity by pushing downward at an angle of 31 ∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26.
To find the force applied by the factory worker to push the crate, we need to consider the forces acting on the crate.
First, we can calculate the vertical force component (Fv) exerted by the worker. This force opposes the weight of the crate and can be calculated using the formula:
Fv = mg
where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s²). Plugging in the given values:
Fv = (29.2 kg)(9.8 m/s²)
Fv ≈ 286.16 N
Next, we can calculate the horizontal force component (Fh) exerted by the worker. This force is responsible for overcoming the frictional force and can be determined using the formula:
Fh = Fk
where Fk is the kinetic friction force. The frictional force is given by:
Fk = μk N
where μk is the coefficient of kinetic friction and N is the normal force. Since the crate is on a level surface, the normal force is equal to the vertical force component (Fv).
Therefore:
Fh = μk Fv
Plugging in the given values:
Fh = (0.25)(286.16 N)
Fh ≈ 71.54 N
The total force applied by the factory worker is the resultant of both the vertical and horizontal force components. Using the angle provided (28° below the horizontal), we can find the magnitude of the applied force (F):
F = √(Fh² + Fv²)
F = √((71.54 N)² + (286.16 N)²)
F ≈ 297.81 N
So, the factory worker pushes the crate with a force of approximately 297.81 Newtons.