Work

Energy

Power Work

done Cases where

work is

not done Positive

and

Negative Work Examples

on

Work Done Work with

Variable

Force Work by

Variable

Force Q&A

Energy

Power Work

done Cases where

work is

not done Positive

and

Negative Work Examples

on

Work Done Work with

Variable

Force Work by

Variable

Force Q&A

See a collection of **solved examples** for this topic on our website
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__Example 1__

A box is dragged horizontally across a floor by a 100 N force acting parallel to the floor. What is the work done by the force in moving it through a distance of 8 m

F = 100 N

S = 8 m

Since F and S are in the same direction,

θ = 0, where θ is the angle of the force to the direction of movement.

W = F Cos θ

= 100 x 8 x Cos 0

= 800 J (Cos 0 = 1)

A box is dragged across a floor by a
100N force directed 60^{o} above the horizontal. How much work
does the force do in pulling the object 8m?

F = 100N

θ = 60^{o}

S = 8m

W = (F Cos θ) S

=(100 Cos 60^{o}) 8

= 100x1/2x 8 = 400 J

A horizontal force F pulls a 10 kg carton across the floor at constant speed. If the coefficient of sliding friction between the carton and the floor is 0.30, how much work is done by F in moving the carton by 5m?

The carton moves with constant speed. Thus, the carton is in horizontal equilibrium.

We look at all the forces acting on the carton:

F is the applied force to the right.

P is the frictional force to the left.

W is the weight of the carton acting downwards.

N is the normal reaction by the floor on the carton acting upwards.

Since there is no movement in the vertical direction W = N = mg

Also since the carton is moving with constant speed,

F = f = μN = μmg.

Thus F = 0.3 x 10 x 9.8

= 29.4 N

Therefore work done W = FS

=(29.4 Cos 0^{o})

=147 J