52.(52G) A 110-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.30. What is the final speed of the crate?

7. ( 11E ) a) Find the momentum of an automobile of mass 2630 kg traveling 21.0 ms .b) Find the velocity ( in km/h ) of a light auto of mass 1170 kg so that it has the same momentum as the auto in part ( a ).

From a to b

Wg= Fx cos 90 = 0
Wn= Fx cos 90 = 0
Wnet=Efk - Eik. Which is = to 1/2mVf^2 -1/2 mVi^2.
Fxcos(0)= Efk. Cos of 0 =1. Vi =0 so use Efk
Fx=1/2mVf^2
2Fx/m = Vf^2
Vf^2= square root of2Fx/m
Vf= 9.77m/s
X = distance, m= mass, F= force, V= velocity
Efk = final kinetic energy and is 1/2 mv^2
Eik= initial kinetic energy
Fn= normal force, Fg, force gravity
B to C
Fg=0
Fn= 0
Wnet = Efk - Eik
Fxcos(0) + Ffrxcos(180). Cos of zero =1 cos of 180 =-1
Fr= 0.30(9.8)(110)(15)(-1). =-4851
Fx - 4851 = Efk - Eik
350 (15)-4851 = Efk - Eik
399= 1/2mVc^2 - 1/2 m Vb^2
399 = 1/2(110)Vc^2 -1/2(110)(9.8)^2
399 = 1/2(110)Vc^2 -5282
399-5282 = 55 Vc^2
399 + 5282/55 = Vc^2
Vc = square root of 399 + 5282/55
Vc=10.2m/s

To find the final speed of the crate, we can break down the problem into two parts:

1. The distance where the floor is frictionless (15 m).
2. The distance where the coefficient of friction is 0.30 (15 m).

Let's first analyze the part where the floor is frictionless. When the floor is frictionless, the only force acting on the crate is the horizontal force of 350 N. This force will accelerate the crate according to Newton's second law, which states that acceleration is equal to the net force divided by the mass of the object.

Given:
- Net force (F) = 350 N
- Mass (m) = 110 kg
- Initial velocity (u) = 0 (starting from rest)
- Distance (d) = 15 m

We can use the equation:
Final velocity squared (v^2) = Initial velocity squared (u^2) + 2 * acceleration (a) * distance (d)

Since the initial velocity is 0 (u = 0), the equation simplifies to:
v^2 = (2 * a * d)

Now, let's calculate the acceleration (a):
Using Newton's second law:
a = F / m
a = 350 N / 110 kg
a = 3.18 m/s^2

Plugging in the values:
v^2 = (2 * 3.18 m/s^2 * 15 m)
v^2 = 95.4 m^2/s^2

Taking the square root of both sides:
v = √(95.4 m^2/s^2)
v = 9.77 m/s

So, the final speed of the crate when the floor is frictionless is 9.77 m/s.

Now, let's calculate the part where the coefficient of friction is 0.30 (15 m). In this case, we have to consider the frictional force.

The frictional force (Ff) can be calculated using the equation:
Ff = coefficient of friction (μ) * normal force (N)

The normal force (N) can be determined using the formula:
N = mass (m) * gravitational acceleration (g)

Given:
- Coefficient of friction (μ) = 0.30
- Distance (d) = 15 m

First, let's calculate the normal force (N):
N = m * g
N = 110 kg * 9.8 m/s^2
N = 1078 N

Next, let's calculate the frictional force (Ff):
Ff = μ * N
Ff = 0.30 * 1078 N
Ff = 323.4 N

Now, let's calculate the net force (Fn):
Fn = Force (F) - Frictional force (Ff)
Fn = 350 N - 323.4 N
Fn = 26.6 N

The net force will cause a deceleration since it opposes the motion. We can calculate the acceleration (a) using Newton's second law:

a = Fn / m
a = 26.6 N / 110 kg
a = 0.242 m/s^2

Now, let's find the final speed using the same equation:
v^2 = (2 * a * d)
v^2 = (2 * 0.242 m/s^2 * 15 m)
v^2 = 7.26 m^2/s^2

Taking the square root of both sides:
v = √(7.26 m^2/s^2)
v = 2.70 m/s

Therefore, the final speed of the crate is 2.70 m/s when the floor has a coefficient of friction of 0.30.

Note: Since the part where the floor is frictionless comes first, it has a greater effect on the final speed.

find the final speed at the end of the first 15 m...

vf^2=2ad where a=force/mass
solve for Vf. Then use that as Vi here

vf^2=vi^2+2ad where a= (F-mu*110g)/mass
and d is again 15m
solve for vf.