Answer: c. -4.84 × 10^5J
Therefore, the work is done by the friction froce against the wheels is -4.84 × 10^5J
Step-by-step 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