Answer to Question #87065 in Mechanics | Relativity for BUDHI PRAKASH

Question

An object of mass 20 kg is pulled up with a force of 150 N on an inclined plane for a
distance of 3.0 m. The plane makes an angle of 30º with the horizontal and the
coefficient of kinetic friction between the object and the plane is 0.3. Draw the free
body diagram and calculate the work done by each force. (Take g = 10 ms−2
)

Answer 1

So,

$$N=m\cdot g\cdot\sin(\alpha)=20\cdot9.81\cdot\sin(30°)=169.91 N$$

$$F_x=-mg\cdot\sin(\alpha)=-20\cdot 9.81\cdot\sin(30°)=-98 N$$

$$F_f=\mu N=0.3\cdot169.91=50.97 N$$

We have

$$A_F=150\cdot 3=450 J$$

$$A_{F_f}=-50.97\cdot 3=-152.9 J$$

$$A_{mg}=A_{mg\cdot \sin(\alpha)}=-98\cdot3=-294.3J$$

$$A_N=0 J$$

$$A_{mg\cos(\alpha)}=0J$$