A block accelerates at 3.3 m/s^2 down a plane inclined at an angle of 26 degrees.Find micro(k) between the block and the inclined plane. The acceleration of gravity is 9.81 m/s^2.
This is my work but i am not sure if it is correct:
sum of forces on x axis= -kinetic friction + mass(gravity) sin theta= mass ( acceleration)x
kinetic friction=fk
sum of forces on y axis=normal force - mg cos theta=ma y
ma y=0
-fk=coefficient of kinetic friction (normal force)
-coefficient(n) + mg sin theta =ma x
-l(-coefficientkf)=(ma x -mg sin theta/n)-1
coeffictientkf=-ma x + mg sin theta/-n
n= mg cos theta
coeffictientkf = -ma x + mgsin theta/-(mg cos theta)
-a x + g sin theta/-g cos theta <-- the masses canceled out
-3.3+9.81 sin 26/ -9.81 cos 26
= -3.694 <-- that cannot be a negative number right?
It looks like you have a good idea of how to solve this; unfortunately, your notation makes it very difficult to follow.
Once you find the normal force (n), you will be working only along the x-axis, so just drop any x-subscripts, so the acceleration is simply a.
Let's let u = the coefficient of kinetic friction, and let T = the angle theta.
I get the equation,
m g sin T - u m g cos T = m a
Simplifying:
u g cos T = g sin T - a
Solve for u.
It looks like you somehow made a "g" -9.81 instead of +9.81.
Remember, g -- as an acceleration -- is a vector; it has direction as well as magnitude. I suspect you introduced a minus sign in an early equation (thus indicating the direction of the vector) and then later used a negative value for g, but magnitudes are always positive.
Your work is mostly correct, but there is a slight mistake in the sign of the friction force. The friction force should have a positive sign, as it opposes the motion down the inclined plane. Let's go through the calculations again:
Forces in the x-direction (parallel to the incline):
Sum of forces = -fk + mgsin(theta) = max
Forces in the y-direction (perpendicular to the incline):
Sum of forces = mgcos(theta) = 0 (since the block is not moving up or down in the y-direction)
Now, let's solve for the friction force sing the equation from the x-direction:
fk = mgsin(theta) - max
Next, we can substitute the given values into the equation:
fk = (mass)(gravity)(sin(theta)) - (mass)(acceleration)
fk = (mass)(9.81)(sin(26)) - (mass)(3.3)
At this point, we can simplify further:
fk = 9.81(mass)(sin(26)) - 3.3(mass)
Now, let's divide both sides of the equation by the normal force (which is equal to the weight of the block, mg):
fk/(mass)(gravity)(cos(theta)) = (9.81(mass)(sin(26)) - 3.3(mass)) / (mass)(gravity)(cos(26))
fk/(mass)(gravity)(cos(theta)) = (9.81(sin(26)) - 3.3) / (gravity)(cos(26))
Finally, we can simplify further:
fk = [(9.81(sin(26)) - 3.3) / (gravity)(cos(26))] * (mass)(gravity)(cos(theta))
Now, substitute the given values for gravity (9.81 m/s^2), the angle theta (26 degrees), and the mass of the block to get the coefficient of kinetic friction:
fk = [(9.81(sin(26)) - 3.3) / (9.81)(cos(26))] * (mass)(9.81)(cos(theta))
Simplified, this becomes:
fk = (9.81(sin(26)) - 3.3) / (cos(26))
Now, plug in the values and calculate the answer:
fk = (9.81(sin(26)) - 3.3) / (cos(26))
≈ 0.443
Therefore, the coefficient of kinetic friction between the block and the inclined plane is approximately 0.443. Note that the coefficient of friction cannot be negative, so your previous answer of -3.694 was incorrect.