Question

A 5.0-kg box rests on a horizontal surface. The coefficient of kinetic friction between the box and surface is μK=0.50. A horizontal force pulls the box at constant velocity for 10 cm. Find the work done by (a) the applied horizontal force, (b) the frictional force, and (c) the net force.

Answer 1

Step-by-step explanation:

Given

mass of box
`m=5 kg`

coefficient of kinetic Friction
`\mu _k=0.5`

Box is moving with constant velocity therefore net Force on object is zero

Force applied=Frictional Force

Force applied
`=\mu mg`

`Force=0.5* 5* 9.8=24.5 N`

(a)work done by applied Force

`W=F\cdot s`

`W=Fs\cos 0=24.5* 0.1=2.45 J`

(b)Work done by Friction force

`W=f\cdot s`

`W=fs\cos (180)`

`W=-fs=-24.5* 0.1=-2.45 J`

(c)Total Work done =work done by Force+Work done by Frictional Force

`W_(net)=2.45-2.45=0 J`