Question

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

Answer 1

Answer:

Work done by the frictional force is $3.41\times 10^5\ J$

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.

$W=k_f-k_i$

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)$

W = −340605 J

or

$W=3.41\times 10^5\ J$

Hence, the correct option is (a).

Answer 2

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

$E=3.4*10^5 J$

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

$E=\frac{1}{2}*mv^2\\\\Therefore\\\\E=\frac{1}{2}*mv^2\\\\E=\frac{1}{2}*(1000)(26.1)^2$

$E=3.4*10^5 J$

Therefore

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

$E=3.4*10^5 J$

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