When the net force acting on an object is zero, the net work done by all the
forces acting on the object is zero. When the net force acting on an object is
not zero, then the net work done on the object is W_{net} =
**F**_{net}**·d**. When a net force acts on an object, then the object accelerates, it changes
its velocity.

Assume an object is moving along a straight line and a constant force
**F**_{net
}= m**a** is acting on the object. Then W_{net} = m**a·d**. The work is
proportional to the acceleration **a**. This acceleration causes a
change in speed, **a**_{ }= (**v**_{f }-
**v**_{i})/∆t.
For translational motion we have

W_{net} = md** **(v_{f }- v_{i})/∆t.

The distance traveled is the average speed times ∆t,

d = ∆t (v_{f }+ v_{i})/2.

Therefore

W_{net} = m(v_{f }- v_{i})(v_{f }+ v_{i})/2 = ½m(v_{f}^{2
}- v_{i}^{2}).

We can express the net work done on the object in terms of the change in the
quantity ½mv^{2}. We define the (translational)** kinetic energy** of the object as K = ½mv^{2}.
The net work done on the object is equal to the change in the kinetic energy of
the object.

W_{net }= K_{f }- K_{i }=
½m(v_{f}^{2}- v_{i}^{2}) = ∆K.

This is called the **work-kinetic energy theorem**.

Assume that together with your partner you want to win a soapbox race. You are allowed to push the cart with your partner in it for a distance of 5 m to give it some initial speed. You are pushing as hard as you can. You do work on the cart. The work you do is the average force you exert times the distance the cart moves in the direction of the force, W = F*(5 m). You transfer energy to the cart. The cart gains kinetic energy.

Kinetic energy increases with the square of the speed. Neglecting friction, an engine does four times as much work to make a car reach a speed of 60 miles/h as to make it reach a speed of 30 miles/h. When the speed of a car is doubled, its kinetic energy increases by a factor of four.

A 3 kg mass has an initial velocity v_{0 }=
6.325 m/s in the positive x-direction.

(a) What is its kinetic energy at this time?

(b) Find the total work done on the object if its velocity changes to 8.944
m/s in the positive x-direction.

Solution:

(a) K = ½mv^{2}, v^{2 }= 40 (m/s)^{2}, K
= ½(3 kg)40
(m/s)^{2 }= 60 J.

(b) The new kinetic energy is ½(3 kg)(8.944)^{2}(m/s)^{2
}= ½(3 kg)80
(m/s)^{2 }= 120 J.

W = 120 J - 60 J =
60 J.

Find the average force needed to bring a 1000 kg car to rest from a speed of 100 km/h in a distance of 130 m.

Solution:

The net work done on the care is W_{net
}= K_{f }- K_{i
}=
½m(v_{f}^{2
}- v_{i}^{2}) = -½m v_{i}^{2}, since v_{f}
= 0.

v_{i} = (100*10^{3} m)/(3600 s) = 27.78 m/s.

Therefore W_{net
}= -½(1000 kg)(27.78 m/s)^{2} =
-3.86*10^{5} J.

A force acts on the car in a direction opposite its direction of travel to
slow it down in a distance of 130 m.

W_{net} = F_{net}*d = F_{net}* (130 m) = -3.86*10^{5}
J.

F_{net} = -2.97*10^{3} N *(1 lbf)/(4.448 N) = 684 lbf.