Q.11 if coefficient of friction between the block and the plane (inclined at 30) is 0.6, then the acceleration of block will be
Answers
1.weight mg ,vertically downward .
2. Normal force ,N perpendicular to the surface of incline.
3. Frictional force , parallel (upward) to incline)
Since we are interested in the motion parallel to the inclined surface, we take one component of mg parallel ( downward) to inclined plane. This component is
mg sin(theta).
Another component of mg perpendicular to inclined surface mg cos(theta) = N, because there is no motion perpendicular to the inclined surface.
Now frictional force f=uN=umg cos (theta). Here, u is coefficient of friction.
If a is acceleration of the body parallel to incline, the equation of motion ( Newton’s second Law of motion) will be
ma= mg sin (theta)- u mg cos (theta). m gets cancelled from both sides. Therefore .
a=g sin (theta)-u g cos (theta).
Put ( theta)= 30 degree , u=0.4 and g=9.8 m/s^2 in this equation. Then,
a=(9.8)(1/2)-(0.4)(9.8)(root(3)/2). I hope now you will find value of a.