# Conservation of Energy problems [ numerical problems solved]

In this post, we will list and solve a set of Conservation of Energy problems. These are basically numerical problems that need to be solved using the principle of Conservation of Energy.

## Conservation of Energy problems | Law of conservation of energy example problems with solutions

Knowing the formulae for various types of energy, we can use them together with the principle of **conservation of energy** to do some calculations. Example problems with solutions are given below.

1] A ball of mass 100 g is thrown vertically upwards with a speed of 12 m/s.

Calculate the maximum height it will reach. g =10 m/s^{2}

**Solution:**

As the ball rises, its **kinetic energy** store will be transferred to a **gravitational potential energy** store. So we can use equations for those quantities to solve the problem.

Potential energy = mgh

Kinetic energy = (1/2) mv^{2}

As energy is conserved, we can write: mgh = (1/2) mv^{2}

=> h = v^{2} /(2g)

So, h =12^{2} / (2×10) = 7.2 m

So, the maximum height = 7.2 m

2] A car, with a mass of 1200 kg and traveling at 30 m/s, is slowed by its brakes to a speed of 20 m/s. The brakes are applied for a distance of 75 m.

Calculate the force the brakes apply in slowing the car down.

**Solution:**

We solve this by thinking of energy transfers.

The braking force acts against the motion of the car and causes retardation. This results in a drop in the kinetic energy of the car.

Thus the work done by the braking force equals the drop in the kinetic energy of the car.

Work done by the braking force = braking force x distance traveled by car = F x 75 = 75F ……….. [1]

Change in KE of the car = (1/2) mv^{2} – (1/2) mu^{2} = (1/2) m[v^{2} – u^{2}] = (1/2) x 1200 x [30^{2} – 20^{2}] = 600 x 500 = 300000 —————- [2]

So, 75F = 300000

=> F = 4000N