A block with a mass of 4.00 kg is placed on an inclined plane inclined at an angle of 35o. The coefficient of kinetic friction between the block and the incline is 0.115. A rope pulls the block up the incline with a force of 32.0 N.
What is the force of friction on the block?
What is the acceleration of the block up the incline?
What is the acceleration of the block down the incline if the rope breaks?
Can you please include a free body diagram?
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 traction force, `F_T,` acts upwards along the inclined plane;
3) the reaction force, `N,` act upwards perpendicular to the inclined plane;
4) the friction force, `F_f,` acts downwards along the inclined plane.
We know `F_f=mu*N,` where mu is a 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, `a` is also along this plane.
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:
So `a=F_T/m - g*(mu*cos(alpha)+sin(alpha))`
which is `approx 32/4-9.8*(0.115*0.82+0.57) approx 8-6.5=1.5 (m/s^2).`
This is the answer to the second question, what is the acceleration of the block up the incline.
The friction force is `mu*m*g*cos(alpha) approx 3.7(N).`
This is the answer to the first question.
If the rope breaks, the equation will be (`F_T=0` , `N` and `F_f` remain the same)
`a= - g*(mu*cos(alpha)+sin(alpha)) approx -6.5(m/s^2),`
or 6.5m/s^2 down the incline. This is the answer to the third question.
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