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

Answer:

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

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

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

$W=Fd\cos (\Theta )$

$W=175 \times 6 \times \cos (20 )J=987J$

Thus work done by the tension force is 987 J

(b)

Normal force on the crate is given by

$N= mg-F\sin \Theta =(40\times 9.8-175\times \sin 20)Newton$

=>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

$\therefore \mu_k=\frac{F\cos (\Theta )}{N}$

=>$\mu_k=\frac{175\times \cos 20}{332}$

=>$\mu _k= 0.495$

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