a block is released from an inclined plane of inclination 45 degree.

Question

find out the acceleration of the block if coefficient of friction between ground and block is 0.5.

Answer 1

To solve we have to resolve forces parallel to and perpendicular to the inclined plane:

 

Parallel:

 

F (component of gravity down the plane) = mgsin45  = 9.81*0.707*m = 6.94m

 

Perpendicular:  

 

N (normal force) = mgcos45 = 9.81*0.707*m = 6.94m

 

f (frictional force opposing motion) = uN = 0.5*6.94m = 3.47m

 

therefore resultant force down the plane = F - f = 6.94m - 3.47m = 3.47m

 

Using Newtons law; force = mass * acceleration;

 

acceleration = force/mass = 3.47m/m = 3.47 m/s²