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 find the speed v and the horizontal force exerted on the car, we can make use of the work-energy principle.
The work-energy principle states that the work done on an object equals the change in its kinetic energy. In this case, the work done on the car is 5,000 J, which is equal to the change in its kinetic energy.
The kinetic energy of an object is given by the formula:
KE = 0.5 * mass * (velocity)^2
Initially, the car is at rest, so its initial kinetic energy is zero. At the final state, the car has a kinetic energy equal to:
KE = 0.5 * mass * (v)^2
Since the work done on the car is equal to the change in its kinetic energy, we can write:
5,000 J = 0.5 * 103 kg * (v)^2
Simplifying the equation:
5,000 J = 51.5 kg * (v)^2
Dividing both sides of the equation by 51.5 kg:
(v)^2 = 5,000 J / 51.5 kg
(v)^2 ≈ 97.09 m^2/s^2
Taking the square root of both sides of the equation, we can find the speed v:
v ≈ √97.09 m^2/s^2
v ≈ 9.85 m/s (rounded to two decimal places)
So, the speed of the car is approximately 9.85 m/s. Now, let's calculate the horizontal force exerted on the car.
The work done on an object is also defined as the product of the force applied and the displacement of the object. In this case, the work done on the car is 5,000 J, and the car moves 21.0 m horizontally.
Therefore, we can write:
5,000 J = force * 21.0 m
To find the force, we divide both sides by 21.0 m:
force = 5,000 J / 21.0 m
force ≈ 238.10 N (rounded to two decimal places)
So, the horizontal force exerted on the car is approximately 238.10 N.
In summary:
(a) The speed v of the car is approximately 9.85 m/s.
(b) The horizontal force exerted on the car is approximately 238.10 N.