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

Explanation

We calculated in the previous chapter that the acceleration experienced by the object in the inclined plane as g `sintheta`.
`" "`According to Newton.s second law, the force acting on the mass along the inclined plane F = mg sin `theta`. 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=vec(F).dvec(r)=Fdrcosphi`
Where `phi` is the angle between the force `(mgsintheta)` and the direction of motion (dr). In this case, force (mg `sintheta`) and the displacement `(dvec(r))` are in the same direction. Hence `phi=0 and cosphi=1`
`W=Fd=(mg sintheta)(dr)`
(dr = length of the inclined place)
`W=1 times 10 times sin(30^(@)) times 10=100 times 1/2=50J`
The component mg `costheta` and the normal force N perpendicular to the direction of motion of the object, so they do not perform any work.