Question

A tension force of 175 N inclined at 20.0° above the horizontal is used to pull a 40.0 - kg packing crate a distance of 6.00 m on a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b) the coefficient of kinetic friction between the crate and surface.

Answer 1

(a) Work done by the tension force is 987 J

(b) Coefficient of kinetic friction between the crate and surface is 0.495

Step-by-step explanation:

(a)

Work done by any force F in moving an object by a distance d making an angle
`\Theta` with the direction of force is given by

`W=Fd\cos (\Theta )`

`W=175 * 6 *  \cos (20 )J=987J`

Thus work done by the tension force is 987 J

(b)

Normal force on the crate is given by

`N= mg-F\sin \Theta  =(40* 9.8-175* \sin 20)Newton`

=>N=332 Newtons

Since crate is moving with constant speed . Therefore using Newtons second law .

`Fcos(\Theta ) -\mu_k N=0`

Where
`\mu_k`=coefficient of kinetic friction

`\therefore \mu_k=(F\cos (\Theta ))/(N)`

=>
`\mu_k=(175* \cos 20)/(332)`

=>
`\mu _k= 0.495`

Thus coefficient of kinetic friction between the crate and surface is 0.495