1.A block accelerates at 3.1 m/s2 down a plane
inclined at an angle 24.0◦.
Find μk between the block and the in-
clined plane. The acceleration of gravity is
9.81 m/s2 .
2.A 1200 kg car moves along a horizontal road
at speed v0 = 20.1 m/s. The road is wet,
so the static friction coefficient between the
tires and the road is only μs = 0.189 and
the kinetic friction coefficient is even lower,
μk = 0.1323.
The acceleration of gravity is 9.8 m/s2 .
What is the shortest possible stopping dis-
tance for the car under such conditions? Use
g = 9.8 m/s2 and neglect the reaction time of
the driver.
Answer in units of m.
Thank you for the help!
wth is this garbage
To solve both of these questions, we will use the concept of friction to determine the required coefficients.
1. In the first question, we are given the acceleration of the block down the inclined plane (3.1 m/s^2) and the angle of the incline (24 degrees). We are required to find the coefficient of kinetic friction (μk) between the block and the inclined plane.
We can start by analyzing the forces acting on the block. There are two main forces at play:
- The gravitational force (down the plane) can be decomposed into two components: the force parallel to the plane (mg*sinθ) and the force perpendicular to the plane (mg*cosθ), where m represents the mass of the block and g is the acceleration due to gravity.
- The frictional force (up the plane) acts in the opposite direction to the motion of the block and opposes its acceleration.
Since the block is moving, we can write the equation for the net force:
Net force = (Mass * Acceleration) - Frictional force
(= m * a - μk * N)
Here, N represents the normal force exerted by the inclined plane on the block, which can be calculated as follows:
N = mg * cosθ
Substituting N into the net force equation, we get:
m * a - μk * mg * cosθ = m * a - mg * sinθ
Now, we can solve for μk:
μk = (m * a - mg * sinθ) / (mg * cosθ)
Plugging in the given values (m = mass of the block, a = acceleration, θ = angle of the incline, g = acceleration due to gravity), we can calculate μk.
2. In the second question, we are given the mass of the car (1200 kg), its initial velocity (v0 = 20.1 m/s), and the coefficients of static friction (μs = 0.189) and kinetic friction (μk = 0.1323). We need to find the shortest possible stopping distance for the car.
To determine the stopping distance, we first need to find the acceleration of the car. We can do this by using Newton's second law:
Net force = Mass * Acceleration
The net force can be calculated as the difference between the driving force (friction) and the resistive force (friction).
- For the static friction (while the car is accelerating), the force is given by the maximum static friction force:
F_s = μs * Mass * g
- Once the car starts skidding, the kinetic friction force comes into play:
F_k = μk * Mass * g
When the car comes to a stop, the kinetic friction force is acting in the opposite direction of the motion. Thus, for the stopping distance, we can set up the following equation:
0 = F_k - F_s
Solving for the stopping distance (d), we get:
d = (v0^2) / (2 * μk * g)
Substituting the given values (v0 = initial velocity, μk = kinetic friction coefficient, g = acceleration due to gravity), we can calculate the shortest possible stopping distance for the car.
Remember to always double-check all calculations and units to ensure accuracy.
2. M*g = 1200*9.8 = 11,760 N. = Wt. of car. = Normal force(Fn).
Fp = Mg*sin 0 = 0.
Fk = uk*Fn = 0.1323*11,760 = 1556 N.
Fp-Fk = M*a.
0-1556 = 1200a, a = -1.30 m/s^2.
V^2 = Vo^2 + 2a*d.
0 = (20.1)^2 - 19.6d, d = ?.
Fp = Force parallel to the incline.
Fn = Normal force.
Divide by Mg:
0.407-0.914u = a/g = 3.1/-9.81,
0.407-.914u = -0.316, uk = 0.791.
1. Fp = Mg*sin24 = 0.407Mg.
Fn = Mg*Cos24 = 0.914Mg.
Fk = u*0.914Mg.
Fp-Fk = M*a.
0.407Mg-u*0.914Mg = M*a.