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A 2.2 × 103 kg car accelerates from rest under the action of two forces. One is a forward force of 1143 N provided by traction between the wheels and the road. The other is a 909 N. Calculate the kinetic energy K of the mass as it reaches 7 m. Answer in units of J.

A 2 2 103 kg car accelerates from rest under the action of two forces One is a forward force of 1143 N provided by

Answer :

When the car reaches 7 m it has a kinetic energy of 14364 J and a terminal velocity of 4.67 m/s.

What is speed?

Speed defines the direction in which a body or object is moving. Speed is primarily a scalar quantity. Velocity is basically a vector quantity. Rate of change of distance. Rate of change of displacement.

What is velocity and its SI units?

(a) The rate of change of displacement is known as velocity. It can be calculated by finding the ratio of displacement and total required time. The SI unit for velocity is "m/s", just like velocity.

The work done by the forward force is:

W1 = F1d = 1143 N × 7 m = 8001 J The work done by the second force is:

W2 = F2d = 909 N x 7 m = 6363 J The net work done on the car is the sum of the following he two works.

[tex]\mathrm{W_{net}}[/tex] = W1 + W2 = 8001 J + 6363 J = 14364 J According to the work energy principle, the net work done in a car is equal to the change in its kinetic energy.

[tex]\mathrm{W_{net}}[/tex] = ΔK where ΔK is the change in kinetic energy. Since the car starts from a standing start, the initial kinetic energy is 0, so ΔK equals the final kinetic energy. So it looks like this:

[tex]\mathrm{W_{net}}[/tex] = 14364 J Finally, we can convert this energy to Joules using the following formula:

1 J = 1 kg m²/s²

So it looks like this:

K = 14364 J = (2.2 × 10³ kg) × v² where v is the final velocity of the car. Solving for v gives:

v = √(K / (2.2 × 10³ kg))

= √(14364 J / (2.2 × 10³ kg))

= 4.67 m/s

So when the car reaches 7 m it has a kinetic energy of 14364 J and a terminal velocity of 4.67 m/s.

To know more about kinetic energy visit:

brainly.com/question/12550364

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