A box of mass 5.0 kg starting from rest slides down a 1.8 m long ramp inclined at an angle of 22.0 deg. from the horizontal. If the ramp is frictionless, how long will it take the box, starting from rest, to reach the bottom of the ramp? if the frictionless ramp is replaced with a ramp having friction between the box and ramp. The time for the box to reach the bottom of the ramp is observed to double. Find the coefficient of kinetic friction between the box and the new ramp.

1. PE=KE

PE=mgh=mg•s•sinα
KE=mv²/2
mg•s•sinα=mv²/2
v=sqrt(2g•s•sinα) =
=sqrt(2•9.8•1.8•sin22°)=3.64 m/s
v=at => a=v/t
s=at²/2= vt²/2t=vt/2
t=2s/v=2•1.8/3.64 =0.99 s
2.
t₁=2t=0.99•2=1.98 s
v₁=2s/t₁=2•1.8/1.98= 1.82 m/s

PE=KE+W(fr)
W(fr) = F(fr)s=μ•mg•cosα•s
mg•s•sinα= mv₁²/2 + μ•mg•cosα•s
μ=(g•s•sinα- v₁²/2)/ g•cosα•s=
=tanα - (v₁²/2g•cosα•s)=
=tan22°-(1.82²/2•9.8•cos22°•1.8)=0.3

To find the time it takes for the box to reach the bottom of the ramp, we can use the principles of physics. Let's first consider the case where the ramp is frictionless.

When the ramp is frictionless, only the force of gravity is acting on the box. We can resolve the weight of the box into two components: one parallel to the ramp and one perpendicular to the ramp.

The component parallel to the ramp is given by:
Force_parallel = m * g * sin(theta),
where m is the mass of the box (5.0 kg), g is the acceleration due to gravity (9.8 m/s²), and theta is the angle of the ramp (22.0 degrees).

Now, we can calculate the acceleration of the box along the ramp using Newton's second law:
a = Force_parallel / m.

Next, we can use the equation of motion for an object sliding down a inclined plane starting from rest:
s = (1/2) * a * t²,
where s is the distance traveled along the ramp (1.8 m) and t is the time taken.

Rearranging the equation and solving for t, we have:
t = sqrt(2 * s / a).

Now, substitute the known values into the equation:
t = sqrt(2 * 1.8 / ((5.0 * 9.8 * sin(22.0))^2)).

Calculating the value, we find:
t ≈ 0.661 seconds.

Now, let's consider the case where the frictionless ramp is replaced with a ramp having friction between the box and the ramp. In this case, the presence of friction will affect the box's motion and increase the time it takes to reach the bottom of the ramp.

We are given that the time taken in this case is double the time taken with a frictionless ramp, so the new time is 2 times the previous time.

Now, let's denote the coefficient of kinetic friction between the box and the new ramp as μ.

The net force acting on the box can be expressed as:
Net_force = m * g * sin(theta) - μ * m * g * cos(theta).

Using Newton's second law as before:
a = Net_force / m.

Next, we can use the equation of motion as before, but with the new acceleration:
s = (1/2) * a * t².

Rearranging the equation and solving for t, we have:
t = sqrt(2 * s / a).

Substituting the known values into the equation and replacing t with 2t (as given), we can solve for the coefficient of kinetic friction μ:

2t = sqrt(2 * 1.8 / ((5.0 * 9.8 * sin(22.0) - μ * 5.0 * 9.8 * cos(22.0))^2)).

Simplifying the equation and solving for μ:

4t² = 2 * 1.8 / ((5.0 * 9.8 * sin(22.0) - μ * 5.0 * 9.8 * cos(22.0))^2).

μ * 5.0 * 9.8 * cos(22.0) = ((2 * 1.8) / (4t²))^0.5 - 5.0 * 9.8 * sin(22.0).

Finally, we can isolate μ and calculate its value using the above equation and the known values.

To find the time it takes for the box to reach the bottom of the frictionless ramp, we can use the equation for the displacement of an object on an inclined plane:

s = ut + (1/2)at^2

Where:
s = displacement of the box on the ramp (1.8 m)
u = initial velocity of the box (0 m/s)
a = acceleration of the box on the ramp (which we need to find)
t = time taken (which we need to find)

We can use the component of the gravitational force parallel to the ramp to find the acceleration:

F_parallel = mgsinθ

Where:
m = mass of the box (5.0 kg)
g = acceleration due to gravity (9.8 m/s^2)
θ = angle of inclination of the ramp (22.0 deg)

So, a = F_parallel / m = (mgsinθ) / m = gsinθ

Plugging in the values:

a = 9.8 * sin(22.0) = 3.82 m/s^2

Now, we can rearrange the equation for displacement to solve for time:

s = ut + (1/2)at^2

1.8 = 0t + (1/2)(3.82)t^2

1.8 = 1.91t^2

t^2 = 1.8 / 1.91

t^2 = 0.9424

t = sqrt(0.9424) = 0.971 s (approximately)

Therefore, it will take approximately 0.971 seconds for the box to reach the bottom of the frictionless ramp.

To find the coefficient of kinetic friction between the box and the new ramp, we can use the fact that the time taken doubles when friction is present.

Let μ be the coefficient of kinetic friction.

The acceleration of the box on the ramp with friction can be determined by subtracting the frictional force from the component of gravitational force parallel to the ramp:

a_with_friction = gsinθ - μgcosθ

According to the problem, the time taken with friction is 2 times the time taken without friction. So, we can set up the following equation:

2t = (1/2)(1.8 / (gsinθ - μgcosθ))

Plugging in the known values:

2(0.971) = (1/2)(1.8 / (9.8sin(22.0) - μ(9.8cos(22.0))))

1.942 = 0.9 / (3.82 - 9.8μ)

Cross-multiplying:

1.942(3.82 - 9.8μ) = 0.9

Expanding:

7.42444 - 18.9772μ = 0.9

Rearranging:

18.9772μ = 7.42444 - 0.9

18.9772μ = 6.52444

μ = 6.52444 / 18.9772

μ ≈ 0.34

Therefore, the coefficient of kinetic friction between the box and the new ramp is approximately 0.34.