Question

Two cars, A and B, are traveling with the same speed of 40 m/sec, each having started from rest. Car A has a mass of 1,200 kg, and car B has a mass of 2,000 kg. Compared to the work required to bring car A up to speed, how much additional work is required to bring car B up to speed?

Answer 1

Answer:

Explanation:

Given

mass of car A $m_a=1200\ kg$

mass of car B $m_b=2000\ kg$

both car start from rest i.e. initial velocity is zero

Final velocity $v=40\ m/s$

Work done required for car A to gain $v=40\ m/s$

$W_a=\frac{1}{2}m_av^2=\frac{1}{2}\times 1200\times 40^2=960000\ J$

$W_a=960\ kJ$

Work done required for car B to gain $v=40\ m/s$

$W_b=\frac{1}{2}m_bv^2=\frac{1}{2}\times 2000\times 40^2=1,600,000\ J$

$W_b=1600\ kJ$

Therefore an additional work of 1600-960=640 kJ is required to bring car B to 40 m/s

Answer 2

The work done on an object will change kinetic energy.

The additional work is required to bring car B up to speed of 40 m/sec is 640 kJ.

What is the work done?

The work done on an object will change kinetic energy. The relation between the work done and kinetic energy is,

$W=KE_i-KE_f$

If the initial velocity is zero then the work done is equal to the kinetic energy. Thus,

$W=KE_f\\W=\dfrac{1}{2}\times m\times v^2$

Here $m$ is the mass of the object and $v$ is the velocity (final) of the object.

Given information-

The speed of the car A and B is 40 m/sec.

The mass of car A is 1200 kg.

The mass of car B is 2000 kg.

The work done by car A to get the speed of 40 m/sec is,

$W_a=\dfrac{1}{2}\times 1200\times 40^2\\W_a=960 \rm kJ.$

The work done by car B to get the speed of 40 m/sec is,

$W_b=\dfrac{1}{2}\times 1600\times 40^2\\W_b=1600 \rm kJ.$

The additional work is required to bring car B up to speed is,

$W_n=W_b-W-a\\W-n=1600-960\\W_n=640 kJ$

Hence, the additional work is required to bring car B up to speed is 640 kJ.

Learn more about the work done and kinetic energy here;

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