An object of mass `m=1 kg` is sliding from top to bottom in the frictionless inclined plane of inclination angle `theta = 30^(@)` and the length of inclined plane is 10m as shown in the figure. Calculate the work done by gravitational force and normal force on the object Assume acceleration due to gravity, `g=10ms^(-2)`

Explanation

We calculated in the previous chapter that the acceleration experienced by the object in the inclined plane as `gsintheta`
According to Newton.s second law, the force acting on the mass along the inclined plane `F=mgsintheta` . Note that this force is constant throughout the motion of the mass The work done by the parallel component of gravitational force `(mg sintheta)` is given by `w=vecF.dvecr=Fdrcosphi`
where is the angle between the force `(mgsintheta)` and the direction of motion `(dvecr)`. In this case, force `(mgsintheta)` and the displacement `(dvecr)` are in the same direction. Hence and `cosphi = 1`
`wFdr=(mgsintheta)(dr)` (dr=lenght of the inclined plane)
`w=1xx10xxsin(30^(@))xx10=100xx1/2=50J` The component ing cos 6 and the normal force N are perpendicular to the direction of motion of the object, so they do not perform any work.