A mass of 1.0 kg is initially held in place on a ramp of angle 45o at height of 20. meters above the ground. The mass is released and slides to the bottom of the ramp (h=0). There is friction between the mass and the ramp. At the bottom of the ramp, the object has a speed of 10. m/s. The thermal energy was generated by the friction as the mass slides down the ramp is closest to which value?

The thermal energy generated by the friction as the mass slides down the ramp is .

Explanation:

Given:

The mass of the object is,

The angle of the ramp is,

The initial height of the object on the ramp is,

The final velocity of the object is,

When the object is at rest on the ramp, its total energy is potential energy. When it moves down the ramp its kinetic energy is increased and potential energy is decreased and a part of its energy is lost to overcome the force of friction. Finally, when it is at the bottom of the ramp, its total energy becomes only kinetic energy.

The total energy of the object at a height on the ramp is given by

When the object is at the bottom of the ramp, its total energy is given by

So, the energy that is lost as thermal energy is given by

double the distance = divide strength by 4

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s