A 2.6 kg block slides down a 25° inclined plane with constant acceleration. The block starts from rest at the top. At the bottom, its velocity is 0.75 m/s. The incline is 1.6 m long.
(a) What is the acceleration of the block?
(b) What is the coefficient of friction?
a. Wb = M*g = 2.6kg * 9.8N/kg=25.48 N. =
Weight of block.
Fp = 25.48*sin25 = 10.77 N. = Force
parallel to the inclined plane.
Fn = 25.48*cos25 = 23.09 N. = Normal Force = Force perpendicular to the
incline.
a=V^2-Vo^2)/2d
a = (0.75^2-0)/2.6 = 0.2163 m/s^2.
b. Fp-Fk = m*a
10.77-Fk = 2.6 * 0.2163 = 0.5625
Fk = 10.77 - 0.5625 = 10.21 N. = Force
of kinetic friction.
u = Fk/Fn = 10.21/23.09 = 0.4421
Correction:
a = (0.75^2-0)/3.2 = 0.1758 m/s^2.
b. Fp-Fk = m*a
10.77-Fk = 2.6*0.1758 = 0.4570
Fk = 10.77 - 0.4570 = 10.31 N
u = Fk/Fn = 10.31/23.09 = 0.4466
To solve this problem, we can use the following equations of motion:
1. For the acceleration of the block down the incline (a):
a = (vf^2 - vi^2) / (2 * d)
2. For the velocity at the bottom of the incline (vf):
vf = vi + a * t
3. For the initial velocity (vi):
vi = 0, because the block starts from rest
4. For the distance traveled down the incline (d):
d = length of the incline = 1.6 m
Using these equations, we can solve the problem step-by-step.
Step 1: Calculate the acceleration (a).
We are given:
vi = 0 m/s (initial velocity)
vf = 0.75 m/s (final velocity at the bottom)
d = 1.6 m (distance down the incline)
Using equation 2, we have:
vf = vi + a * t
Since the block starts from rest (vi = 0), the equation simplifies to:
vf = 0 + a * t
a * t = vf
a = vf / t
Step 2: Calculate the time (t).
To find the time, we can use the equation of motion for distance:
d = vi * t + (1/2) * a * t^2
Since the block starts from rest (vi = 0), the equation simplifies to:
d = (1/2) * a * t^2
t^2 = (2 * d) / a
t = sqrt((2 * d) / a)
Step 3: Substituting values and solving for acceleration (a).
Substituting the values we are given:
vf = 0.75 m/s
d = 1.6 m
From Step 1, we know that a = vf / t. Substituting the value of t from Step 2:
a = vf / sqrt((2 * d) / a)
a^2 = vf^2 / (2 * d)
a = sqrt(vf^2 / (2 * d))
Step 4: Calculate the coefficient of friction (μ).
The coefficient of friction can be found using the equation:
μ = tan(θ)
Given that the angle of the incline (θ) is 25°, we can substitute it to find μ:
μ = tan(25°)
Now let's calculate the values:
a) Acceleration (a):
a = sqrt(0.75^2 / (2 * 1.6))
a ≈ 0.501 m/s^2
b) Coefficient of friction (μ):
μ = tan(25°)
μ ≈ 0.466
To find the acceleration of the block, we can use the following equation:
a = (vf^2 - vi^2) / (2d)
Where:
a = acceleration
vf = final velocity
vi = initial velocity (which is 0 since the block starts from rest)
d = distance traveled along the inclined plane
In this case, the final velocity (vf) is 0.75 m/s and the distance traveled (d) is 1.6 m. Plugging these values into the equation, we get:
a = (0.75^2 - 0^2) / (2 * 1.6)
Simplifying this expression, we get:
a = 0.5625 / 3.2
a ≈ 0.176 m/s^2
So, the acceleration of the block is approximately 0.176 m/s^2.
Now, to find the coefficient of friction (μ), we can use the following equation:
μ = tan(angle)
Where:
angle = angle of the inclined plane
In this case, the angle of the inclined plane is given as 25°. Taking the tangent of this angle, we get:
μ = tan(25°)
Using a calculator, we find that:
μ ≈ 0.466
So, the coefficient of friction is approximately 0.466.