Question

A force acts on a 2 Kg object so that its position is given as a function of times as x=3t²+5. What is the work done by this force in the first 5 seconds?

Answer 1

Given parameters are

 

The mass of an object = 2Kg

The function of time: x=3t²+5

 

Force: A force is a push or pull upon an object resulting from the object’s interaction with another object.

 

Newton’s second law, F= ma

Where F is force, m is mass of an object and ‘a’ is acceleration.

 

Differentiating the above equation with respect to time gives velocity

⇒ v = dx/dt

Differentiating x=3t²+5 on both sides with t

dx/dt = 6t+0

v = 6t+0

 

At t = 0

⇒v = 0

 

t = 5 sec

⇒ v = 30 m/s

 

When a force forces a body to move, the force does work on the item. When a force (F) moves an item across a distance, it is measured in work (d). When work is completed, energy is moved from one energy storage to another, resulting in the equation: energy transferred= work completed.

 

The S.I unit of work is joule (J).

Work done = ΔK.E

Where KE is the Kinetic Energy.

Work done = 1/2mv²−0

=12(2)(30)²

= 900 J

 

 

Work done = 900J