A factory worker pushes a 30.1 kg crate a distance of 5.0 m along a level floor

Question

A factory worker pushes a 30.1 kg crate a distance of 5.0 m along a level floor at constant velocity by pushing downward at an angle of 29 degrees below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.

1)What magnitude of force must the worker apply to move the crate at constant velocity?

2)How much work is done on the crate by this force when the crate is pushed a distance of 5.0 ?

3)How much work is done on the crate by friction during this displacement?

4)How much work is done by the normal force?

5)How much work is done by gravity?

6)What is the total work done on the crate?

Explanation / Answer

N=294.98 N Let F be the force applied by worker F cos 29 - u(N+Fsin29) =0 F(0.75826) = 0.24x294.98 1)F=93.3646 N 2)Fcos29 x 5 = 408.2929 J 3)Work done by friction is same as friction = Fcos 29 therefore W=408.2929 J 4) Work done by norma is zero . 5) Work done by gravity is Zero. 6) no net work done as Force acting on the crate is zero.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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