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## Main Question or Discussion Point

A car of mass 1kg rests on top of a large flat mass of 1,000,000 kg (10^6 kg). They are at rest (relative to the sun) in space. The car has a battery and an electric motor with 100% efficiency. We can assume that when a torque is applied to the wheels, they roll on the surface of the flat mass without any slippage or rolling resistance.

The engine applied a torque to the wheel for 1 second in such a way that the force on the car is exactly 1N.

m1 = 1 kg, m2 = 10^6 kg, dt = 1s, F = 1 N.

Using F = ma, we get:

F = m1 * a1;

a1 = F/m1 = 1/1;

a1 = 1 N/s^2 (acceleration of car). Since dt = 1 s, we get:

v1 = 1 m/s;

Similarly, we find that that

V2 = -0.000001 m/s = -10^-6 m/s.

The principle of conservation of momentum gives the same result.

Now the car is travelling along the flat mass with a velocity of 1m/s (relative to the sun) and a relative velocity of 1.000001m/s relative to the flat mass.

The kinetic energy of the car relative to the sun is KE1 = ½ * m1*v1^2.

KE1 = ½*1*1*1 = 0.5 J.

The kinetic energy of the flat relative to the sun is KE2 = ½ * m2*v2^2.

KE1 = ½*10^6*10^-6*10^-6 = 5*10^-7 J.

Total KE is 0.5000005 J.

All this energy came from the battery of the car and all of it got converted to kinetic energy.

At this point the engine applied another torque to the wheel for 1 second second in such a way that the force on the car is exactly 1N.

Can we assume that the amount of energy required if almost the same for the second period of applied force as it was for the first? Why?

The engine applied a torque to the wheel for 1 second in such a way that the force on the car is exactly 1N.

m1 = 1 kg, m2 = 10^6 kg, dt = 1s, F = 1 N.

Using F = ma, we get:

F = m1 * a1;

a1 = F/m1 = 1/1;

a1 = 1 N/s^2 (acceleration of car). Since dt = 1 s, we get:

v1 = 1 m/s;

Similarly, we find that that

V2 = -0.000001 m/s = -10^-6 m/s.

The principle of conservation of momentum gives the same result.

Now the car is travelling along the flat mass with a velocity of 1m/s (relative to the sun) and a relative velocity of 1.000001m/s relative to the flat mass.

The kinetic energy of the car relative to the sun is KE1 = ½ * m1*v1^2.

KE1 = ½*1*1*1 = 0.5 J.

The kinetic energy of the flat relative to the sun is KE2 = ½ * m2*v2^2.

KE1 = ½*10^6*10^-6*10^-6 = 5*10^-7 J.

Total KE is 0.5000005 J.

All this energy came from the battery of the car and all of it got converted to kinetic energy.

At this point the engine applied another torque to the wheel for 1 second second in such a way that the force on the car is exactly 1N.

Can we assume that the amount of energy required if almost the same for the second period of applied force as it was for the first? Why?