A factory worker pushes a 30.0kg crate a distance of 4.7m along a level floor at

Question

A factory worker pushes a 30.0kg crate a distance of 4.7m along a level floor at constant velocity by pushing downward at an angle of 30? below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26.

A: What magnitude of force must the worker apply to move the crate at constant velocity?

B:How much work is done on the crate by this force when the crate is pushed a distance of 4.7m ?

C: How much work is done on the crate by friction during this displacement?

D: How much work is done by the normal force?

E: How much work is done by gravity?

F: What is the total work done on the crate?

Explanation / Answer

On box in Vertical direction,

Fnet = N - mg - Fsin30 = 0

N = mg + Fsin30

In horizontal,

Fnet = Fcos30 - f = 0

friction force f = u.N = 0.26 ( mg + Fsin30 )

putting this value in above equation,

Fcos30 = 0.26 ( mg + Fsin30 )

F (cos30 - 0.26sin30) = 0.26mg

F = 103.96 N   ..........Ans(A)

B)   Wok done = F.d = Fdcos30 = 103.96 x 4.7 x cos30 = 423.17 J

C) Friction force = f = Fcos30 = 90.03 N

Work done = f.d = fdcos180 = 90.03 x 4.7 x cos180 = -423.17 J

D) Work done by normal W = N.d = Ndcos90 = 0

E) Work done by gravity = mg. d = mgdcos90 = 0  

F) total work done = sum of work by all forces

= 423.17 + (-423.17) + 0 + 0 = 0 J

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