A block with a mass of 3.00 kg is placed on an inclined plane inclined at an angle of 26o. The inclined plane is 1.20 m long and the block takes 1.10 s to get to the end of the inclined plane.
What is the acceleration of the block?
What is the force of friction on the block?
What is the coefficient of friction between the surfaces?
Can you please include a free body diagram if possible?
1 Answer | Add Yours
Denote the angle of inclination as `alpha.`
Let's draw a free-body diagram. The forces acting on our block are:
1) the gravity force, `mg,` acts down;
2) the reaction force, `N,` acts upwards perpendicular to the inclined plane;
3) the friction force, `F_f,` acts upwards along the inclined plane.
We know `F_(f)=mu*N,` where `mu` is the coefficient of friction.
Together they make the net force `F` which is equal to `ma` where `a` is the acceleration of the block (by Newton's Second Law). Because the block moves along the inclined plane, the acceleration is also along this plane.
Acceleration `a` is a constant (because all forces are constant), and it takes 1.10s to go though 1.20m. The equation of the movement along the incline is
and `S(1.10)=0,` so `1.20=(a*(1.1)^2)/2,` or `a = 2.4/(1.1)^2 approx 1.98 (m/s^2).`
This is the answer for the first question.
Now please look at the picture. We'll use x-axis parallel to the inclined plane and y-axis perpendicular to it.
Consider projection on the y-axis:
And on the x-axis:
Substitute `F_(f)=mu*N=mu*m*g*cos(alpha)` into the second equation:
Therefore `mu=(a+g*sin(alpha))/(g*cos(alpha)) approx (1.98+9.8*0.44)/(9.8*0.90) approx 0.71.`
This is the answer to the third question.
The last, `F_(f)=mu*N=mu*m*g*cos(alpha) approx 0.71*3*9.8*0.9 approx 18.79 (N).`
This is the answer to the second question.
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