2. A wooden block slides directly down an inclined plane, at a constant velocity of 6 m/s.
a. How large is the coefficient of kinetic friction if the plane makes an angle of 25 degrees with the horizontal?
b. If the angle of incline is changed to 10º, how long far will the block slide before coming to a stop?
What is tangentTheta?
forcedownplane=frictionupplane
mg*sinTheta=mu*mg*cosTheta
sintheta/costheat=mu
tangent theta=mu
b. net force down=ma
mg*sinTheta-mu*mgCosTheta=ma
solve for acceleration a. Mu is tangent of 25.
normal force = m g cos 25
friction force = mu m g cos 25
= force down slope = m g sin 25
so
sin 25/cos 25 = tan 25 = mu
= .466
===================
at 10 degrees
friction force = .466 m g cos 10
= .459 m g
work done by friction
= .459 m g d
= loss of kinetic energy
= (1/2)m(36) = 18 m
so
18 = .459 g d = .459*9.81 * d
d = 4 meters
oh and for time avergae speed = 6/2 = 3 m/s
so
t = 4/3 seconds
a. To find the coefficient of kinetic friction, we need to use the formula:
\(F_{friction} = \mu_k \cdot F_{normal}\)
Where:
\(F_{friction}\) is the force of friction,
\(\mu_k\) is the coefficient of kinetic friction, and
\(F_{normal}\) is the normal force.
Since the block is sliding directly down the inclined plane, the normal force is equal in magnitude and opposite in direction to the component of the weight perpendicular to the inclined plane (known as the normal force). The equation for this normal force is:
\(F_{normal} = m \cdot g \cdot \cos(\theta)\)
Where:
\(m\) is the mass of the block,
\(g\) is the acceleration due to gravity (approximately 9.8 m/s²), and
\(\theta\) is the angle of inclination (25 degrees in this case).
Now, since the block is sliding down the inclined plane at a constant velocity of 6 m/s, we know that the net force acting on the block is zero. Since the force of gravity can be split into two components, parallel to the incline and perpendicular to the incline, the force equation becomes:
\(F_{friction} = m \cdot g \cdot \sin(\theta)\)
Setting this equal to the kinetic friction equation, we get:
\(m \cdot g \cdot \sin(\theta) = \mu_k \cdot m \cdot g \cdot \cos(\theta)\)
Cancelling out the mass and acceleration due to gravity:
\(\mu_k = \tan(\theta)\)
Substituting the value of the angle, we have:
\(\mu_k = \tan(25°)\)
Using a scientific calculator, evaluate \(\tan(25°)\), which is approximately 0.4663.
Therefore, the coefficient of kinetic friction is approximately 0.4663.
b. To find how far the block will slide before coming to a stop, we need to use the equation:
\(v^2 = u^2 + 2as\)
Where:
\(v\) is the final velocity (zero in this case as it comes to a stop),
\(u\) is the initial velocity (6 m/s in this case),
\(a\) is the acceleration,
and \(s\) is the distance.
Since the block is sliding up the inclined plane, \(a\) can be calculated using:
\(a = -g \cdot \sin(\theta)\)
Where:
\(g\) is the acceleration due to gravity (approximately 9.8 m/s²),
and \(\theta\) is the angle of inclination (10 degrees in this case).
Rearranging the formula, we get:
\(s = \frac{v^2 - u^2}{2a}\)
Substituting the values into the formula:
\(s = \frac{0 - (6 \, \text{m/s})^2}{2 \cdot -9.8 \, \text{m/s}^2 \cdot \sin(10°)}\)
Use a scientific calculator to evaluate the expression and calculate \(s\).
Therefore, the block will slide approximately 3.08 meters before coming to a stop.