A 1.92 × 103 kg car accelerates uniformly from rest to 12.4 m/s in 3.77 s.
What is the work done on the car in this time interval?
a=(V-Vo)/t = (12.4-0)/3.77 = 3.29 m/s^2
d = 0.5*3.29*3.77^2 = 23.37 m.
Work = F*d = (m*g) * d=(1920*9.8)23.37 =
439,805 Joules.
To find the work done on the car, we can use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy.
The formula for work is:
Work = Force * Distance * cos(θ)
Given that the car is accelerating uniformly, we can use the equation for uniform acceleration kinematics:
v = u + at
where v is the final velocity, u is the initial velocity (0 in this case since the car starts from rest), a is the acceleration, and t is the time interval.
In this case, the final velocity, v, is 12.4 m/s and the time interval, t, is 3.77 s. We need to find the acceleration, a.
Rearranging the equation for velocity, we get:
v = u + at
12.4 = 0 + a * 3.77
Solving for a, we divide both sides of the equation by 3.77:
a = 12.4 / 3.77
Now that we have the acceleration, we can calculate the work done on the car using the formula:
Work = Force * Distance * cos(θ)
Since the car is moving in a straight line, the force is equal to the mass of the car multiplied by the acceleration:
Force = mass * acceleration
Mass = 1.92 × 10^3 kg (given)
Work = (1.92 × 10^3 kg)(12.4 / 3.77) * Distance * cos(θ)
The distance is not given in the problem, so we cannot determine the exact work done on the car without that information. The work done will depend on the distance over which the car accelerates.