A factory worker pushes a 29.6 kg crate a distance of 4.6 m along a level floor

Question

A factory worker pushes a 29.6 kg crate a distance of 4.6 m along a level floor at constant velocity by pushing downward at an angle of 32 Degree below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25. What magnitude of force must the worker apply to move the crate at constant velocity? Express your answer using two significant figures. How much work is done on the crate by this force when the crate is pushed a distance of 4.6 m? Express your answer using two significant figures.

Explanation / Answer

given

m = 29.6 kg

d = 4.6m

theta = 32degrees

mue_k = 0.25

Let N is the normal force on the crate.

Workdone by normal force = N*d*cos(theta)

= N*d*cos(90)

= 0 <<<<<<<-------------Answer

Total workdone on crate = Fnet*d

= 0*d (since the crate is moving with constant velocity net force acting on it must be zero)

= 0 <<<<<<<-------------Answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.