A block of mass 4 kg slides down a plane inclined at `37^(@)` with the horizontal. The length of plane is 3 m. find work done by normal force , gravity and frictional force .

Explanation

As the normal reaction is perpendicular to the displacement, therefore
Work done by the normal reaction `N = NS cos 90^(@) = 0`,Work done by the gravity
` = mg S cos(90^(@) - 37^(@) = 53^(@)` (angle between mg and displacement)
`mg S sin 37^(@) = 4 xx 9.8 xx 3 xx 3/5 = 70.56 J`
Work done by friction ` = -(mu N)S = -(mu mg cos 37^(@))S`
(here `-ve` sign indicates that frictional force and dispalcement are in opposite direction) ` = -0.2 xx 4 xx 9.8 xx 4/5 xx 3`
` = -18.816 J`.