Question

Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, the box (initially at rest) is traveling at a speed of 6 m/s. How much work (in Joules) did friction do in this process

Answer 1

Answer: 321 J

Step-by-step explanation:

Given

Mass of the box
`m=3\ kg`

Force applied is
`F=25\ N`

Displacement of the box is
`s=15\ m`

Velocity acquired by the box is
`v=6\ m/s`

acceleration associated with it is
`a=(F)/(m)`

`\Rightarrow a=(25)/(3)\ m/s^2`

Work done by force is
`W=F\cdot s`

`W=25* 15\\W=375\ J`

change in kinetic energy is
`\Delta K`

`\Rightarrow \Delta K=(1)/(2)m(v^2-0)\\\\\Rightarrow \Delta K=(1)/(2)* 3* 6^2\\\\\Rightarrow \Delta K=(1)/(2)* 3* 36\\\\\Rightarrow \Delta K=54\ J`

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

`\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J`

Therefore, the magnitude of work done by friction is
`321\ J`