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).

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!

it's a b and