A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn), the kinetic frictional force is (fk), and the weight of the box is (mg). Decide which forces do positivem zero, or negative work.

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

FOR FORCE (P):

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

Therefore, work done by force (P) is Positive.

FOR NORMAL FORCE (Fn) AND WEIGHT (W):

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.

FOR KINETIC FRICTIONAL FORCE (fk):

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

Therefore, work done by Kinetic Frictional Force (fk) is Negative.

the bal will not change its velocity.

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