A driver traveling at 22m/s, slows down her 2000kg car to stop for a red light. What work is done by the friction froce against the wheels?

a. - 2.2X104 J

b. - 4.4X104 J

c. - 4.84X105 J

d. - 9.68X105 J

Answer: c. - 4.84 * 10^5J

Therefore, the work is done by the friction froce against the wheels is - 4.84 * 10^5J

Explanation:

Given;

Car velocity v = 22m/s

Mass m = 2000kg

From the law of conservation of energy.

Kinetic energy of the car before stopping K. E = workdone by friction force against the wheels W

W = - K. E ... 1

And the kinetic energy of the car can be written as

K. E = 1/2 mv^2 ... 2

Substituting m and v into equation 2

K. E = 1/2 * 2000kg * 22^2

K. E = 484,000J

From eqn1

W = - K. E = - 484,000J

W = - 4.84 * 10^5J

Therefore, the work is done by the friction froce against the wheels is - 4.84 * 10^5J

C - 4.84 * 10^5J

Explanation:

Work done = force * distance

v^2 = u^ + 2as

u = 22m/s

a = 10m/s^2

When the car stops the final velocity (v) = 0

0 = 22^2 + 2*10*s

s = - 484/20

s = -24.2m

Work done = force * distance

Force = mass * acceleration

Work done = 2000*10 * - 24.2

= - 4.84*10^5J