A 1.53×10

3
kg car accelerates uniformly from
rest to 13.6 m/s in 2.44 s.
What is the work done on the car in this
time interval?
Answer in units of J

Correction:

a = (V-Vo)/t = 13.6-0)/2.44=5.57 m/s^2.

F = m*a = 1530*5.57 = 8522.1 N.

d=0.5*a*t^2 = 0.5*5.57*(2.440^2=16.6 m.

Work = F*d = 8522.1*16.6 = 141467 Joules

To find the work done on the car, we can use the work-energy principle:

Work = Change in Kinetic Energy

The change in kinetic energy can be found using the equation:

ΔKE = 1/2 * m * (v_f^2 - v_i^2)

Where:
ΔKE is the change in kinetic energy
m is the mass of the car
v_f is the final velocity of the car
v_i is the initial velocity of the car (which is 0, since it starts from rest)

Given:
m = 1.53×10^3 kg
v_f = 13.6 m/s
v_i = 0

Now, let's calculate the work done on the car:

ΔKE = 1/2 * (1.53×10^3 kg) * (13.6 m/s)^2
= 1/2 * 1.53×10^3 * 185.76
= 141,940.16 J

Therefore, the work done on the car in this time interval is approximately 141,940.16 J.

Wc = m*g = 1530kg * 9.8N/kg = 14,994 N.

= Wt. of car.

a = (V-Vo)/t = (13.6-0)/2.44=5.57 m/s^2.

d = 0.5a*t^2 = 0.5*5.57*(2.44)^2=16.6 m.

Work = F*d = 14994*16.5 = 248780 Joules.