A mechanic pushes a 2.50 x 10^3 kg car from the rest to a speed of v, doing 5000J of work in the process. During this time, the car moves 25.0 m. Neglecting friction between car and road, find (a) v and (b) the horizontal force exerted on the car.

(a) 2.00 m/s (b)200 N

Can you please show the work

To find the answers, we'll make use of the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy.

(a) To find the final speed of the car (v), we need to consider the work done on the car and its change in kinetic energy.

According to the work-energy principle, the work done on the car (W) is equal to the change in kinetic energy (ΔKE), which is given by:

W = ΔKE

Given that the work done on the car is 5000 J, we can say:

5000 J = ΔKE

The change in kinetic energy (ΔKE) is equal to the final kinetic energy (KEf) minus the initial kinetic energy (KEi) of the car. Since the car is initially at rest, the initial kinetic energy (KEi) is zero.

ΔKE = KEf - KEi
ΔKE = KEf - 0
ΔKE = KEf

Therefore, we can rewrite the equation as:

5000 J = KEf

Using the equation for kinetic energy:

KE = 0.5 * m * v^2

where m is the mass of the car, we can rewrite the equation as:

5000 J = 0.5 * (2.50 x 10^3 kg) * v^2

Now, solve for v:

v^2 = (5000 J) / ((0.5 * 2.50 x 10^3 kg))
v^2 = (5000 J) / (1250 kg)
v^2 = 4 m^2/s^2

To find v, take the square root of both sides:

v = √(4 m^2/s^2)
v = 2 m/s

So, the final speed of the car (v) is 2 m/s.

(b) To find the horizontal force exerted on the car, we can use the formula for work:

Work = Force * Distance

The work done on the car is given as 5000 J, and the distance the car moves is 25.0 m. Therefore, we have:

5000 J = Force * 25.0 m

Now, solve for the force (F):

Force = 5000 J / 25.0 m
Force = 200 N

So, the horizontal force exerted on the car is 200 N.

To find the speed (v) and the horizontal force exerted on the car, we can utilize the work-energy principle. According to this principle, the work done on an object is equal to the change in its kinetic energy.

(a) To find the speed (v), we can solve for it using the work done (W) and the change in kinetic energy (∆KE). The work done on the car is equal to the change in kinetic energy.

Given:
Work done (W) = 5000 J
Mass of the car (m) = 2.50 x 10^3 kg
Distance moved by the car (d) = 25.0 m

First, let's find the change in kinetic energy (∆KE). The initial kinetic energy (KEi) is zero since the car starts from rest, and the final kinetic energy (KEf) is (1/2)mv^2, where v is the speed.

∆KE = KEf - KEi
∆KE = (1/2)mv^2 - 0
∆KE = (1/2)mv^2

Since ∆KE = Work done (W), we can set up the equation:

(1/2)mv^2 = W

Plugging in the given values:
(1/2)(2.50 x 10^3 kg)(v^2) = 5000 J

Simplifying the equation:
(v^2) = (2 x 5000 J) / (2.50 x 10^3 kg)
(v^2) = 10000 J / (2.50 x 10^3 kg)
(v^2) = 4 J/kg
v^2 = 4 J/kg
v = √(4 J/kg)

Taking the square root:
v ≈ 2 m/s

Therefore, the speed of the car (v) is approximately 2 m/s.

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

Since the car is moving horizontally and the net force is horizontal, the acceleration is also horizontal. Thus, the net force can be written as:

Fnet = m * a

Using Newton's second law,

Fnet = m * a

The car's acceleration (a) can be determined using the equation:

v^2 = u^2 + 2as

Where u is the initial velocity (0 m/s), s is the distance moved by the car (25.0 m), and v is the final velocity (2 m/s).

v^2 = u^2 + 2as
(2 m/s)^2 = (0 m/s)^2 + 2*a*(25.0 m)
4 m^2/s^2 = 50a
a = 4 m^2/s^2 / 50
a = 0.08 m/s^2

Now we can calculate the net force (Fnet):

Fnet = m * a
Fnet = (2.50 x 10^3 kg) * (0.08 m/s^2)
Fnet = 200 N

Therefore, the horizontal force exerted on the car (Fnet) is 200 N.