Suppose an object is pushed up a smooth inclined plane from ground level to a point h above the ground with force F. The angle of inclination is A. How much work is done?
To push an object up a smooth inclined plane, one must overcome the force of gravity which causes the object to slide down the plane. Hence, the force F=mgsinA must be applied. We can denote the distance the body undergoes as s=hsinA, then:
W=Fs=mg(sinA)^2
The answer is mgh and I know that this can be easily proven using energy conservation laws, but I want to know why my method is incorrect.
Best Answer
You've made a mistake in calculating the displacement component for the object. If you observe, $S\neq hsinA$
$$S=\frac{h}{\sin A}$$
and therefore, the work done, is:
$$F.s=\left(mg\sin A\right)\left(\frac{h}{\sin A}\right)=mgh$$
which is the exact expression you got from energy conservation.