A mechanic pushes a 3.6 multiplied by 103 kg car from rest to a speed of v, doing 4600 J of work in the process. During this time, the car moves 27.0 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
N

(a) The final kinetic energy,

(M/2)(Vfinal)^2,
is the work done.
(b) Since the work done is also
F*X = F*27.0 m,
you can solve for the horizontal force, F.

To find the answers to the given questions, we can use the work-energy principle. According to this principle, the work done on an object is equal to its change in kinetic energy.

Given:
Mass of the car, m = 3.6 x 10^3 kg
Work done, W = 4600 J
Distance traveled, d = 27.0 m

(a) To find the speed (v) of the car, we need to calculate the change in kinetic energy.

The work done on an object is equal to the change in its kinetic energy, so we can write:

W = ΔKE

Since the car starts from rest, its initial kinetic energy (KEi) is zero. Therefore, the change in kinetic energy can be written as:

W = KEf - KEi

Substituting the given values, we get:

4600 J = KEf - 0

Therefore, the final kinetic energy (KEf) is equal to 4600 J.

We know that kinetic energy (KE) is given by the formula:

KE = (1/2)mv^2

Substituting the known values, we have:

(1/2)(3.6 x 10^3 kg)(v^2) = 4600 J

Now, we can solve for v:

1.8 x 10^3(v^2) = 4600 J

v^2 = 4600 J / (1.8 x 10^3 kg)

v^2 = 2.56 m^2/s^2

Now, taking the square root of both sides, we get:

v = √(2.56 m^2/s^2)

So the speed (v) of the car is approximately 1.6 m/s.

(b) To find the horizontal force exerted on the car, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration.

The acceleration of the car can be calculated using the following formula:

v^2 = u^2 + 2as

where:
v is the final velocity (speed) of the car
u is the initial velocity (which is zero in this case)
a is the acceleration of the car
s is the distance traveled by the car

Substituting the known values, we have:

(1.6 m/s)^2 = 0^2 + 2a(27.0 m)

2.56 m^2/s^2 = 54a

a = 2.56 m^2/s^2 / 54

a ≈ 0.047 m/s^2

Now, we can calculate the force (F) using Newton's second law:

F = ma

Substituting the known values, we get:

F = (3.6 x 10^3 kg)(0.047 m/s^2)

F ≈ 169.2 N

Therefore, the horizontal force exerted on the car is approximately 169.2 N.