The work done by a ____ force pushing a duster on a frictionless surface by 0.25 m is 12.5 J

Concept:

When a force is applied on an object which causes displacement along the direction of force then the work is said to be done by that force.

The work done by a force is defined as: The product of the component of the applied force on the block in the direction of the displacement and the magnitude of this displacement is called work done.

i.e., \(W = \vec F.d\vec x = Fdx\cos \theta \)

or in terms of magnitude, it can be given as

\(W\; = \;F \times s\)

Here both force and displacement will be along the same direction, since force is causing this displacement thus Work done is a scalar quantity.

Here,

Force applied on object = F

Displacement of object = dx

Distance traveled by an object = s

The angle between Force and displacement = θ

Calculation:

Given that,

Distance duster pushed, s = 0.25 m

Work was done to displace this duster, W = 12.5 J

Now from the above expression the force acting on a duster can be given as

\(W = F \times s \Rightarrow F = \frac{W}{s} = \frac{{12.5}}{{0.25}} = 50\;N\;\)

Thus option 4 is correct