An object of mass 10.0 kg is released from the top of an inclined plane which makes an angle of inclination of 30.0o with the horizontal. the object slides along the inclined plane. the questions refer to the instant when the object has traveled through a distance of 2.00 m measured along the slope. the coefficient of kinetic friction between the mass and the surface is 0.200. use g = 10.0 m/s2. how much work is done by gravity?

We define work as the net force multiplied with the distance traveled. We already know distance to be 2 meters. Thus, all we need to know is the net force.

When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are three forces acting on the box: gravitational force pointing straight down, normal force perpendicular to the slope denoted as Fn and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
           Fn = (10)(10)(cos 30) = 86.6 N
∑Fx = μk*Fn - mgsin30° = Fnet
          (0.2)(86.6 N) - (10)(sin30°) = Fnet
Fnet = 12.32 Newtons

Thus,
W = Fnet*d
W = (12.32 N)(2 meters)\
W = 24.64 Joules

The objects do not reach terminal velocity         there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.

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