A factory worker pushes a 28.1 k g crate a distance of 5.0 m along a level floor

Question

A factory worker pushes a 28.1kg crate a distance of 5.0m 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.

What magnitude of force must the worker apply to move the crate at constant velocity? in N

How much work is done on the crate by this force when the crate is pushed a distance of 5.0m ? in J

How much work is done on the crate by friction during this displacement? in J

Explanation / Answer

Let the Force be F. The pushing force F(p) must be equal to the friction force F(f)

Let the friction coeficient be: u. F(f)= u*(m*g+F*sin Q), where Q is the angle below the horisontal, m is the mass of the crate and g is the gravitation.

The Pushing force F(p)=F*cosQ

Therefore: F*cosQ = u*m*g+u*F*sinQ), or F*(cosQ-u*sinQ)=u*m*g, so: F=u*m*g/[cosQ-u*sinQ]
F = 98.995N

b)W = F.dcos31 = 424.27J

c)Wf = -F.d = -424.27J

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