Question

A 4 kg block is placed at the top of an inclined plane. The plane is 2.5 meters long and inclined at 34°. The coefficient of kinetic friction between the block and plane is 0.27. The block slides the 2.0 meters down the ramp. What speed does it have at the bottom?

Answer 1

Work done is given by: W=Fs cos, where F is magnitude of force vector, s is magnitude of displacement vector and is the angle between force and displacement vectors.

Work done by gravity:

Here,force F=mg, where m is mass and g is gravitational acceleration. Here,m=4 kg. So,F=4*9.8=39.2 N

Displacement , s=2 m

=90-34=56 degrees.

So,work done=39.2 *2 *cos56=43.84 J

Work done by friction:

Here,force F=mgcos, where m is mass and g is gravitational acceleration, is coefficient of friction, is angle of ramp with horizontal. Here,m=4 kg,=0.27,=34 degrees. So,F=0.27*4*9.8*cos34=8.7745 N

Displacement , s=2 m

=180 degrees as displacement and frictional force are opposit to each other.

So,work done=8.7745 *2 *cos180= - 17.55 J

So,total work done=43.84-17.55=26.29 J.

Now,according to work energy theorem, work done = change in kinetic energy.

Also, kinetic energy=1/2mv2 where m is mass and v is velocity.

Initially the object is at rest, so initial kinetic energy=0.

So, final kinetic energy-initial kinetic energy=26.29

=>1/2mv2=26.29, where v is the required final velocity.

=>1/2*4*v2=26.29

=>v2=26.29*2/4=13.145

=>v=3.63 m/s.

So,required speed=3.63 m/s