A mechanic pushes a 2.0 103 kg car from rest to a speed of v, doing 5350 J of work in the process. During this time, the car moves 29.0 m. Neglecting friction between car and road, find each of the following.

(a) the speed v
m/s
(b) the horizontal force exerted on the car

for a) w= 1/2 (m)(v^2) so just plug in the work that is given and the mass, and solve for v.

for b) w=fd so f=w/d so plug in work and distance, convert as needed, and you're set. Hope I helped, even though I answered your question almost 3 years late!

To find the answers, 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.

(a) First, we can find the final kinetic energy of the car.
We know that work done is equal to the change in kinetic energy, so 5350 J = (1/2) * m * v^2 - (1/2) * m * 0^2. As the car starts from rest, the initial kinetic energy is zero.

Simplifying the equation, we get:
5350 J = (1/2) * m * v^2

Now we can plug in the values:
m = 2.0 * 10^3 kg (mass of the car)
v = ?

5350 J = (1/2) * (2.0 * 10^3 kg) * v^2

We can solve for v^2:
v^2 = (2 * 5350 J) / (2.0 * 10^3 kg)
v^2 = 5,350 J / 10^3 kg
v^2 = 5.35 m^2/s^2

Finally, taking the square root of both sides gives us:
v ≈ √(5.35) ≈ 2.31 m/s

So, the speed of the car, v, is approximately 2.31 m/s.

(b) The work done on the car is the work done by the horizontal force exerted on it. The work done is given by the equation: work = force * distance.

We know that the work done is 5350 J and the car moves a distance of 29.0 meters. So we can rearrange the equation to solve for force:

force = work / distance
force = 5350 J / 29.0 m

Evaluating the expression gives us:
force ≈ 184.48 N

Therefore, the horizontal force exerted on the car is approximately 184.48 N.