A 70 kg cart is pushed for 50 m with a constant velocity upon a 45º frictionless incline. What is the work done on the cart?
Work is defined as force through a distance:
1) Work = Force * Distance, or W=fd
The cart has a mass of 70 kg, subject to the acceleration of gravity, 9.8 m/sec^2.
Force is defined in terms of Newtons (N) as (kg m)/sec^2, so the cart has a force of
70kg * 9.8 m/sec^2 = 686 (kg m)/sec^2 or 686 N
so by formula 1, the Work should be 686N * 50m, however, there's an incline to consider which modifies the force. Formula 1 is better described as:
2) Work = Force * Distance(cos (theta)), where theta is the angle of the incline. So the formula should read:
Work = 686N * 50m(cos 45)
= 686N * 50m(.7071)
= 686N * 35.355m
= 24253 Nm
A Newton * meter (Nm) is the same thing as a Joule (J), which is used to express an amount of work, so
= 24253 J
The work done w in moving the mass of m kg to a vertical height h meter is given by: w= mgh, where g is the acceleration due to gravity =9.81m/s^2.
Here height of the 50 m inclination of 45 degree = 50 sin 45.
Therefore, W = 70*g*50cos 45 = 70*9.81*50 sin45 = 24278.51133 J