1. What magnitude of force must the worker apply to move the crate at constant v

Question

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 4.3m ?

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?

A factory worker pushes a 28.8kg crate a distance of 4.3m along a level floor at constant velocity by pushing downward at an angle of 31 degree below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.

Explanation / Answer

Mass of the crate m = 28.8 Kg

Acceleration a = 0 ( since velocity is constanat)

Coefficient of kinetic friction ? = 0.25

Let F be the applied force.

Distance covered S = 4.3 m

Friction f = ?[mg + F sin31o]

Here mg = 28.8 kg x 9.8 m/s2 =282.242 N

-----------------------------------------------------------------------------------------

SOLUTION:

(A) From Newtons second law we can write

                       ma = F cos31 - f

                         0 = 0.857 F - 0.25[ 282.242 N + F sin31o]

               0.857 F = 70.56 N + 0.128 F

             0.729 F = 70.56 N

F = 96.79 N

                       

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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