A mechanic pushes a 2.60 103-kg car from rest to a speed of v, doing 5,000 J of work in the process. During this time, the car moves 21.0 m. Neglecting friction between car and road, find each of the following.
(a) the speed v?
(b) the horizontal force exerted on the car?
To answer these questions, we need to use the work-energy principle and the equations of motion.
(a) To find the speed v, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.
The work done on the car is given as 5,000 J. The change in kinetic energy of the car is equal to its final kinetic energy minus its initial kinetic energy.
Since the car starts from rest, its initial kinetic energy is zero. The final kinetic energy can be calculated using the equation:
Kinetic energy = (1/2) * mass * velocity^2
Substituting the known values, we have:
5,000 J = (1/2) * (2.60 * 10^3 kg) * v^2
Solving for v:
v^2 = (2 * 5,000 J) / (2.60 * 10^3 kg)
v^2 = 3,846.15
v ≈ 62.03 m/s
Therefore, the speed of the car is approximately 62.03 m/s.
(b) To find the horizontal force exerted on the car, we can use Newton's second law, which states that the force acting on an object is equal to its mass multiplied by its acceleration.
Since the car is moving horizontally, the net force acting on it is equal to the product of its mass and its acceleration in the horizontal direction. In this case, neglecting friction, the only horizontal force on the car is the force applied by the mechanic.
We can calculate the acceleration of the car using the equation of motion:
v^2 = u^2 + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Rearranging the equation to solve for acceleration, we have:
a = (v^2 - u^2) / (2s)
Since the car starts from rest, the initial velocity (u) is zero. Plugging in the known values:
a = (62.03 m/s)^2 / (2 * 21.0 m)
a = 191.55 m/s^2
Now, we can calculate the horizontal force:
Force = mass * acceleration
Force = (2.60 * 10^3 kg) * 191.55 m/s^2
Force ≈ 4.97 * 10^5 N
Therefore, the horizontal force exerted on the car is approximately 4.97 * 10^5 N.