Question

A block with a mass of 2.5kg lies on a surface that is inclined at an angle of, θ=22o, with the horizontal. The coefficient of friction between the block and the surface of the incline is, uk=0.20.

a) Apply Newton’s 2nd Law to determine the acceleration of the block as it slides down the

surface. Answer -->  [±1.9m/s2]

b) Determine the magnitude of the velocity of the block at the base of the inclined surface

if the block starts from rest and moves through a distance of 3.41m from its initial

position to the base of the incline. Answer --> [3.6m/s]

c) Apply the Work/Energy Theorem to determine the velocity of the block at the base of

the inclined surface if the block starts from rest and moves through a distance of 3.41m from its initial position to the base of the incline Answer --> [3.6m/s]

Answer 1

part a:

component of weight of the block along the incline in downward direction

=mass*g*sin(theta)=9.17786 N

normal force=mass*g*cos(theta)=22.716 N

friction force=friction coefficient*normal force

=4.5432 N

friction force will oppose the motion

hence total force along the inclince in downward direction

=9.17786-4.5432=4.5754 N

then acceleration=force/mass=1.83016 m/s^2

part b:

initial velocity=0

distance travelled=3.41 m

then final velocity=sqrt(initial velocity^2+2*acceleration*distance)

=3.533 m/s

part c:

initial potential energy=mass*g*height

=2.5*9.8*3.41*sin(22)=31.2965 J

work done against friction=force*distance

=4.5432*3.41=15.4923 J

then kinetic energy at the base=initial energy-work done against friction

=15.8042 J

if speed is v,

then 0.5*mass*v^2=15.8042

==>v=sqrt(2*15.8042/2.5)=3.555 m/s