Question

A 3 kg object is moving along a horizontal surface. The kinetic energy of the object is increasing at a constant rate of 6 J/m; that is, the energy increases linearly with the distance travelled. If the object starts from rest how far (in m) has it moved after 4 s?

Answer 1

To solve this problem we will apply the concepts of energy conservation and Newton's second law that defines force as the product of the object's mass with its acceleration. Additionally we will apply concepts related to the kinematics equations of linear motion.

For conservation of energy we have that work is equal to kinetic energy therefore,

$W = KE$

$Fd = \frac{1}{2} mv^2$

Here,

F = Force

d = Displacement

m = Mass

v= Velocity

At the same time we have the relation of

$F = \frac{W}{d}$

Therefore the value of the force can be interpreted as the rate of increase in energy per unit of distance, which makes it equivalent to

$F = \frac{W}{d} = 6J/m$

Applying Newton's Second Law

$F = ma$

$6J/m = (3kg)a$

$a = 2m/s^2$

In 4 seconds final velocity of the object becomes

$v = at$

$v= 2*4$

$v= 8m/s$

Then the work done is equal to,

$W = KE$

$W = \frac{1}{2} mv^2$

$W = \frac{1}{2} (3)(82)$

$W = 96J$

Then the displacement is,

$W = F*d$

$d = \frac{W}{F}$

$d = \frac{96}{6}$

$d = 16 m$

Therefore the distance moved is 16m