A child slides down a rough incline of height 4.0 m and has an angle of 30 degre

Question

A child slides down a rough incline of height 4.0 m and has an angle of 30 degrees with the horizontal. The child has a weight of 40 kg and the coefficient of kinetic friction is .2. What is the child’s velocity at the bottom of the incline assuming the child starts at rest? A child slides down a rough incline of height 4.0 m and has an angle of 30 degrees with the horizontal. The child has a weight of 40 kg and the coefficient of kinetic friction is .2. What is the child’s velocity at the bottom of the incline assuming the child starts at rest?

Explanation / Answer

let speed at the bottom is v m/s.

friction force=friction coefficient*normal force

=0.2*mass*g*cos(theta)

=0.2*40*9.8*cos(30)=67.896 N

using work energy principle,

initial potential energy+initial kinetic energy-work done against friction=final potential energy+final kinetic energy

==>40*9.8*4+0.5*40*0^2-67.896*(4/sin(30))=40*9.8*0+0.5*40*v^2

==>1024.8=20*v^2

==>v=sqrt(1024.8/20)=7.1582 m/s

hence child's speed at the bottom is 7.1582 m/s.

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