potential engergy and kinetic energy acceleration

Question

3.2 A motor car with a mass of 850 kg accelerates uniformly from rest up a gradient of 1 in 40 and reaches a speed of 60 km/h after 4 minutes. Calculate the following: 3.2.1 The acceleration of the motor car 3.2.2 The kinetic energy that the motor car possesses at the end of 4 minutes 3.2.3 The gain in potential energy

Answer 1

3.2.1

The mass of the car is 850 Kg.

Gradient is 1 up 40.

We know that F = ma.

Now look at the diagram

Gradient = 1 to 40 then

Then Force acting along the inclined surface is F = mg sinθ then a = g sinθ

a = 9.8 *0.02499

a = 0.2449 m/sec²

Therefore accerlation of the motor car is a = 0.2449 m/sec²

Answer 2

(3.2.2)

Mass of the motor Car is 850 Kg.

At t = 4 minutes, v = 60km/hr

1 km/hr = 0.2778 m/sec

60 km/hr = 60 * 0.2778 m/sec

V = 16.668 m/sec

Kinetic Energy = (mv²)/2

E = (850*(16.668)²)/2

E =118,057 J

Therefore Kinetic Energy E = 118.05 KJ.

Answer 3

(3.2.3)

Potential Energy is given as

PE = mgh.

We know that where s is the Height of the inclined Surface.

Initial Velocity is Zero and a=g

h = (16.667)² / 2*9.8 = 14.17 mts

At Height h = 14.17.

PE = mgh

E = 850 * 9.8 *14.17

E = 118,060.27 J

The Gain in Potential Energy is given as E = 118.06 KJ.

Therefore It satisfies the law of Conservation of Energy.

At top of the Inclined surface, Potential Energy PE is 118.06 KJ and  Kinetic Energy KE = 0 (since Initial Velocity is Zero).

Total Energy at the top of the surface = PE + KE = 118.06 KJ

At bottom of the Inclined surface, Potential Energy PE is 0 and Kinetic Energy KE = 118.06 KJ (since Height is Zero).

Total Energy at the bottom of the surface = PE + KE = 118.06 KJ.

Therefore Law of conservation of Energy is satisfied.