acceleration kinetic energy

Question

3.2 A truck with a mass of 4 tonnes accelerates uniformly from the rest up a slope of 4% and reaches a speed of 90 km/h after 3 minutes. Neglect friction and calculate the following: 3.2.1 The acceleration of the truck 3.2.2 The kinetic energy that the truck possesses after 3 minutes 3.2.3 The gain in potential energy in MJ

Answer 1

3.2.1)

Mass of the truck is  4 tones  =  4000 kgs 

Slope percentage is 4% (Slope percentage / 100)

To convert slope percentage to degrees we use formula  ø =  tan^-1 (Slope percentage / 100) 

 ø =  tan^-1 (4 / 100) 

 ø =  tan^-1 (0.04)

 ø = 2.29°

So the Ramp is inclined with  2.29° .

The coefficient of kinetic friction (μk)= 0  .                [ neglecting the friction ]

The net force accelerating the sled up the ramp, F_a, is the difference between the component of the sled’s weight along the ramp and the frictional force opposing it:

F_a  =   mg sin ø - μk mg cosø  

F_a  =   mg sin ø                         [ since μk = 0 ]

F_a  =   (4000) (9.8) sin 2.29  

F_a  = 39200*0.0399

F_a  = 1566.33

So the net force is 1566.33 N .

Acceleration of  sled (a) = F_a /m                                 [ since F = ma ]

a = 1566.33 / 4000 

a = 0.391 m/s² .

The acceleration of the truck  is 0.391 m/s² .

 

Answer 2

3.2.2)

The velocity of the truck after 3 min is 90 km/h (v) = 25 m/s .

Mass of the truck is  4 tones  =  4000 kgs  .

So the kinetic energy that the truck possesses after 3 minutes is = 1/2 (mv²) 

KE = 1/2 [(4000) (25²)]

KE = 1/2 [2500000]

KE = 1250000 J

So the kinetic energy that the truck possesses after 3 minutes is 1250000 J .