A box is given a push up a 16.1° incline. When it reaches the bottom again it is only going 0.59 its original speed. Find the coefficient of kinetic friction.
*I know the answer is .06 but I don't know how!
It's an interesting problem.
Do you happen to be given the initial velocity as 0.7 m/s ?
I have an impression that μ is dependent on the initial velocity.
Unfortunately I wasn't given any other information. I'm thinking that you may have to solve in terms of variables and then the variables may cancel out?
Yes, I did it differently, and this time I can eliminate all other variables.
I get 0.06 m/s only if the final speed has been reduced by 0.59 of the original speed. If the final speed is 0.59 of the original speed, then μ=0.1396.
Let:
t=theta=16.1°
g=acceleration due to gravity
h=maximum height reached
μ=coefficient of kinetic friction
v0=initial speed
v1=return speed=0.59v0
Assume gravitational potential energy is zero at the starting (and ending) position.
Assume static friction equals kinetic friction so that box does not get "stuck" at the highest point.
Next we will determine total energy and work done at three instants,
A) at the start,
KE=(1/2)v0^2
PE=0
Work=0
Total energy,
E1 = (1/2)v0²
B) at the highest point
KE=0
PE=mgh
Work=energy dissipated=μmgcos(t)h/sin(t)
Total energy
E2 = mgh+μmgcos(t)h/sin(t)
C) at the end point,
KE=(1/2)v1²
PE=0
Work=energy dissipated=2μmgcos(t)h/sin(t)
Total energy
E3 = (1/2)v1² + 2μmgcos(t)h/sin(t)
Assuming no other energy lost/gained, then
E1=E2
E1=E3=2*mu*m*g*cos(t)*h/sin(t)/(1-0.59^2)
=>
E2=E3 ..............(1)
Solve for μ
m,g,h will all cancel out, giving
μ=0.1396 m/s
Note: if v1=(1-0.59)V0, then μ=0.060823...
To find the coefficient of kinetic friction, we can use the concept of conservation of energy and the work-energy theorem.
First, let's analyze the initial situation:
1. The box is pushed up a 16.1° incline, which means it is moving against the force of gravity.
2. At the highest point, the box comes to a momentary stop before descending down the incline.
3. No external work is done on the box along the incline.
Now, let's consider the final situation:
1. The box reaches the bottom of the incline and is moving with a speed that is 0.59 times its original speed.
2. The box is still moving on a horizontal surface, and the only force acting on it is the force of friction.
The change in kinetic energy from the top to the bottom of the incline is equal to the work done by the net force acting on the box. We can express this mathematically as:
ΔKE = work
The change in kinetic energy can be calculated as:
ΔKE = (1/2) * m * v_f^2 - (1/2) * m * v_i^2
where m is the mass of the box, v_f is the final velocity, and v_i is the initial velocity.
Assuming no loss of energy due to other factors (such as air resistance), the work done by the net force (consists of gravitational force and friction force) can be expressed as:
work = m * g * h - f_k * d
where g is the acceleration due to gravity, h is the vertical height traversed, f_k is the kinetic friction force, and d is the distance covered along the incline.
Since the box returns to its original speed at the bottom of the incline:
v_f = 0.59 * v_i
Now, let's substitute these values into the equations and solve for f_k:
(1/2) * m * (0.59 * v_i)^2 - (1/2) * m * v_i^2 = m * g * h - f_k * d
Simplifying further:
(0.59^2 - 1) * (1/2) * m * v_i^2 = m * g * h - f_k * d
Since h = d * sin(16.1°):
(0.59^2 - 1) * (1/2) * m * v_i^2 = m * g * d * sin(16.1°) - f_k * d
Now, rearrange the equation to solve for f_k:
f_k * d = m * g * d * sin(16.1°) - (0.59^2 - 1) * (1/2) * m * v_i^2
f_k = (m * g * d * sin(16.1°) - (0.59^2 - 1) * (1/2) * m * v_i^2) / d
We can further simplify this equation:
f_k = g * sin(16.1°) - (0.59^2 - 1) * (1/2) * v_i^2
Now, we substitute the known values:
- g = 9.8 m/s^2 (acceleration due to gravity)
- sin(16.1°) = 0.278 (sine of the angle)
f_k = 9.8 * 0.278 - (0.59^2 - 1) * (1/2) * v_i^2
Finally, substituting v_i = v_f / 0.59:
f_k = 9.8 * 0.278 - (0.59^2 - 1) * (1/2) * (v_f / 0.59)^2
Now, plug in the value of v_f / 0.59 (0.59 times the original speed) into the equation and calculate f_k:
f_k = 9.8 * 0.278 - (0.59^2 - 1) * (1/2) * (0.59)^2
Calculating this expression, we get:
f_k ≈ 0.06
Therefore, the coefficient of kinetic friction is approximately 0.06.