Question

A body of mass 10kg slides down on an
inclined plane inclined at 30degrees to the horizontal.if muk=0.5.
find the force acting on it.​

Answer 1

Answer:

Force acting on the body = 6.75 N

Explanation:

Given that,

A body of mass 10kg slides down on an

inclined plane

here,

mass of the body = 10 kg

and given inclination = 30°

and constant of kinetic friction of the inclined = 0.5

since body is sliding

so,

friction will exert upward parallel to the inclined

and force due to its mass = mg

and parallel to inclined = mg sin@

where,

@ = 30°

and normal force(N) exerted by the plane perpendicular to it = mg cos@

and friction = ûN

= û mg cos@

now

force acting on it = downward force - friction

putting the values,

force = mg sin30 - ûmg cos30

= mg(½ - 0.5√3/2)

= 10(10)((1 - 0.5√3)/2)

= 100(1 - 0.5(1.73))/2

= 100(1 - 0.865)/2

= 50(0.135)

= 6.75 N

so,

Force acting on the body = 6.75 N

Answer 2

Answer:

Force acting on the body = 6.75 N

Explanation:

Given,

mass of body = 10 kg

and inclination(@) = 30°

and constant of kinetic friction = 0.5

and force due to its mass parallel to inclined = mg sin@

and normal force(N) exerted by the plane perpendicular to it = mg cos@

and friction = ûN

= û mg cos@

now

force acting on it = downward force - friction

putting the values,

force = mg sin30 - ûmg cos30

= mg(½ - 0.5√3/2)

= 10(10)((1 - 0.5√3)/2)

= 100(1 - 0.5(1.73))/2

= 100(1 - 0.865)/2

= 50(0.135)

= 6.75 N

so,

Force acting on the body = 6.75 N