Question

A 335 kg box is pulled 6.00 m up a 30° frictionless, inclined plane by an external force of 5425 N that acts parallel to the plane.

Calculate the work done by the external force.

Calculate the work done by gravity.

Calculate the work done by the normal force.

Answer 1

Part A.

Along the incline, external force applied = F_app = 5425 N

displacement of box = 6.00 m

So,

Work = F.d = F*d*cos

= Angle between Force and displacement = 0 deg

So,

W_app = F_app*d*cos 0 deg = F_app*d

W_app = 5425*6.00

W_app = 32550 J

Part B.

Component of Force of gravity along the inclined plane will be:

F_g = m*g*sin

= Angle of inclined plane = 30 deg

d = displacement along the incline = 6.00 m

= Angle between Force of gravity (downward) and displacement (upward) = 180 deg

So,

W_g = F_g*d*cos

W_g = 335*9.8*sin 30 deg*6.00*cos 180 deg

W_g = -9849 J

Part C.

Since Normal Force is always perpendicular to displacement, So

W_n = N*d*cos

= Angle between Normal force and displacement = 90 deg

W_n = N*d*cos 90 deg = N*d*0

W_n = 0 J

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