Physics question: A crate of mass 451kg is loaded onto a railway car using a ramp. The ramp is 6.12m long and the bed of the railway car is 1.53m above the ground. A force applied parallel to the ramp moved the crate at a constant speed p up the ramp. Calculate the useful work done by moving the crate up the ramp.
The work developed by a force is defined as the product of the force, the distance and the cosine of the angle that form the direction of the force and the direction of motion.
W = F*d*cosθ
When the crate is on the ramp the weight can be decomposed in a component wy = mgcosθ perpendicular to the ramp and a component wx = mgsinθ parallel to the ramp, where θ is the angle between the ramp and the ground.
To load the crate on the railway car, it is only necessary to apply a force in the direction parallel to the ramp (if friction is not taken into account), which is at least equal to the parallel component of the weight.
Then we can write:
W = (mg sinθ)(d) cos 0°
The sine of the angle can be found by dividing the height of the railway car (opposite side), between the length of the ramp (hypotenuse):
sinθ = (1.53)/(6.12) = 0.25
Substituting in the equation above:
W = (541)(9.8)(0.25)(6.12)
W = 8111.7 J
The useful work done by moving the crate up the ramp is 8111.7 Joules.