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:


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.


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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.


Work done on a object by definition is $W = \mathbf{F}\cdot\mathbf{S}$. This is a vector product so you can project one vector on the another to get the result. At first you were projecting the force $\mathbf{F}$ on the displacement $\mathbf{S}$. You got the answer here and projecting the displacement on the vertical axis and multiply it by the force along the slope is meaningless.


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