A crate of mass 37 kg is pushed up a ramp by a person as shown in the figure bel

Question

A crate of mass 37 kg is pushed up a ramp by a person as shown in the figure below. The person pushes the crate a distance of 23 m as measured along the ramp. Assume the crate moves at constant velocity. Assume the coefficient of kinetic friction between the crate and the ramp is K = 0.37. What is the work done by the person as he pushes the crate up the ramp?

Can you please explain how to do this probleem and what formulas I need to use. Thank you

A crate of mass 37 kg is pushed up a ramp by a person as shown in the figure below. The person pushes the crate a distance of 23 m as measured along the ramp. Assume the crate moves at constant velocity. Assume the coefficient of kinetic friction between the crate and the ramp is ¼K = 0.37. What is the work done by the person as he pushes the crate up the ramp? Can you please explain how to do this probleem and what formulas I need to use. Thank you

Explanation / Answer

Assuming no acceleration along the ramp...
Sum of the forces along the crate = 0 = F - m*g*sin(30) - friction

Sum of the forces perpendicular to the ramp = 0 = N - m*g*cos(30)
N = m*g*cos(30)

Remember, friction = u*N = u*m*g*cos(30)

0 = F - m*g*sin(30) - friction = F - m*g*sin(30) - u*m*g*cos(30)

F = m*g*sin(30) + u*m*g*cos(30)

F = 37*9.8*1/2 + 0.37*37*9.8*0.866 = 297.48 N

Work = F*d = 297.48 * 23 = 6842.04 J

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