Answer:
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
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
