Question

A block of mass 2 kg is released from the top of a rough incline of height 2 m if the velocity of the block on reaching the bottom of the incline is 4 m/s find the magnitude of the work done in joules by the frictional force on block? G=9.8m/s2

Answer 1

`"23.2 J"`

Explanation:

Initial total energy of block is

`"T.E"_i = "mgh = 2 kg × 9.8 m/s"^2 × "2 m = 39.2 J"`

Energy of block when it reaches the bottom is

`"T.E"_f = 1/2\ "mv"^2 = 1/cancel(2) × cancel(2) "kg" × ("4 m/s")^2 = "16 J"`

From conservation of energy

Initial total energy = Final Total energy

Initial total energy = Energy when it reaches the bottom + Energy lost in friction

Energy lost in friction `"= 39.2 J - 16 J = 23.2 J"`

`"∴ 23.2 J"` of work is done by frictional force on block.