How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 j b.2.73 x 10^5 j c.4.09 x 10^5 j d.4.77 x 10^5 j

We have that for the Question "How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest?" it can be said that work-done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest is

From the question we are told

How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest?

a.3.41 x 10^5 J

b.2.73 x 10^5 J

c.4.09 x 10^5 J

d.4.77 x 10^5 J

Generally the equation for the Kinetic Energy  is mathematically given as

Therefore

work-done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest is

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Work done by the frictional force is

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

W = −340605 J

or

Hence, the correct option is (a).

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