An object is pulled across a rough surface with a force of 10 N. If the object moves with a constant velocity and the surface has a coefficient of kinetic frict

Question

An object is pulled across a rough surface with a force of 10 N. If the object moves with a constant velocity and the surface has a coefficient of kinetic friction of 0.2, what is the magnitude of the normal force acting on the object?

Answer 1

Hi Deb,

 

F= 10N horizontal  force applied

Constant velocity implies that the sum of all the forces acting on the body are zero i.e. acceleration =0

Y direction W-N=0 N= normal force so N=W

X direction F-.2N=0 so N=F/.2 =10/.2=50N

 

Hope this helps

 

Jim

 

Answer 2

Remember that friction forces oppose forces that result in motion, static friction keeps objects from sliding down inclined planes and friction force helps moving objects maintain contact with the surface the object is moving on.

 

From Newton's Second Law of motion,

 

F = ma,  a = 0 because the object moves with constant velocity, F is the net force on the object

 

F = 0,  the force of friction is f = μR where R is the reaction force due to the surface and it is equal to normal force on the object 

 

pulling force - friction force = 0

        10 - 0.2*R = 0

        10             = 0.2 R

        10/0.2       = R

 

Therefore the normal force = 50 N