A car whose brakes are locked skids to a stop in 200 ft from an initial velocity of 50 mi/hr. Find the coefficient of sliding friction.
First convert 50 mi/hr to ft/s
Vo= 73.33 ft/s
g = 32.2 ft/s^2
M*Vo^2/2 = M*g*Uk*X
M cancels out
Vo^2 = 2*g*Uk*X = 5378 ft/s^2
Uk = 5378/(2*32.2*200)= 0.4175
thank you!
Well, I have to say, that's one car that really knows how to slam on the brakes! Let's get down to business and solve this problem. To find the coefficient of sliding friction, we can use the following equation:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (0 ft/s, because the car comes to a stop)
vi = initial velocity (50 mi/hr converted to ft/s)
a = acceleration (which is caused by the friction)
d = distance (200 ft)
So, plugging in the values we have:
0^2 = (50 * 5280/3600)^2 + 2 * a * 200
Now, let's solve for the coefficient of sliding friction:
a = -vi^2 / (2d)
a = -((50 * 5280/3600)^2) / (2 * 200)
a ≈ -1.59 ft/s^2
The negative sign shows that the acceleration is in the opposite direction to the motion. Now, we can further calculate the coefficient of sliding friction:
a = μ * g, where g = 32.2 ft/s^2 (acceleration due to gravity) and μ is the coefficient of sliding friction.
So:
μ = a / g
μ ≈ -1.59 ft/s^2 / 32.2 ft/s^2
μ ≈ -0.049
Now, it seems we have encountered a little hiccup here. The negative value for the coefficient of sliding friction suggests that something is off. Since friction is always a positive force, it's possible that there was an error in the calculations or assumptions made. It's best to recheck the calculations and make sure we have all the correct values. Keep in mind that my jokes are accurate, but I may make a mistake from time to time!
To find the coefficient of sliding friction, we can use the equation:
d = (v^2 - u^2) / (2a)
where:
d = distance covered (200 ft)
v = final velocity (0 mi/hr since the car skids to a stop)
u = initial velocity (50 mi/hr)
a = acceleration
Step 1: Convert the initial and final velocities to ft/s.
Since 1 mi/hr is equivalent to 1.47 ft/s, we have:
u = 50 mi/hr * 1.47 ft/s/mi ≈ 73.5 ft/s
v = 0 mi/hr * 1.47 ft/s/mi = 0 ft/s
Step 2: Substitute the given values into the equation.
200 ft = (0^2 - 73.5^2) / (2a)
Step 3: Simplify the equation.
200 ft = (-73.5^2) / (2a)
Multiplying both sides by (2a) gives:
400a ft = -73.5^2
Step 4: Solve for a.
a = -73.5^2 / 400
a ≈ -13.52 ft/s^2
Step 5: Substitute the value of a into the equation for the coefficient of sliding friction.
a = g * μ
where:
g = acceleration due to gravity (32.2 ft/s^2)
μ = coefficient of sliding friction
-13.52 ft/s^2 = 32.2 ft/s^2 * μ
Step 6: Solve for μ.
μ = -13.52 ft/s^2 / 32.2 ft/s^2
μ ≈ -0.420
The approximate coefficient of sliding friction is -0.420. Note that the negative value indicates that the direction of friction opposes the motion of the car.
To find the coefficient of sliding friction, we can use the equation of motion for a skidding car:
d = (v^2 - u^2) / (2a),
where d is the distance traveled, v is the final velocity, u is the initial velocity, and a is the acceleration.
Given:
Initial velocity (u) = 50 mi/hr
Distance traveled (d) = 200 ft
First, we need to convert the initial velocity from miles per hour to feet per second.
1 mile = 5280 feet
1 hour = 3600 seconds
Thus, the initial velocity u = 50 mi/hr is converted to:
u = (50 * 5280) / 3600 ft/s = 73.33 ft/s
Now, we rearrange the formula to solve for the acceleration (a):
a = (v^2 - u^2) / (2d)
Since the car comes to a stop, the final velocity (v) is 0 ft/s.
So, the equation becomes:
0^2 - (73.33)^2 = 2a(200).
Simplifying:
-5374.43 = 400a.
Now, solve for the acceleration (a):
a = -5374.43 / 400 = -13.436 ft/s^2.
The negative sign indicates that the acceleration is in the opposite direction of motion, as the car is decelerating.
Finally, we can use the equation for the coefficient of sliding friction (μ):
μ = (a / g),
where g is the acceleration due to gravity (32.2 ft/s^2).
μ = (-13.436 / 32.2) ≈ -0.417.
Since the coefficient of sliding friction cannot be negative, we take the absolute value of -0.417 to get:
μ ≈ 0.417.
Therefore, the coefficient of sliding friction is approximately 0.417.