m = mass of the person = 82 kg
g = acceleration due to gravity acting on the person = 9.8 m/s²
F = normal force by the surface on the person
f = kinetic frictional force acting on the person by the surface
μ = Coefficient of kinetic friction = 0.45
The normal force by the surface in upward direction balances the weight of the person in down direction , hence
F = mg eq-1
kinetic frictional force on the person acting is given as
f = μ F
using eq-1
f = μ mg
inserting the values
f = (0.45) (82) (9.8)
f = 361.6 N
Answer:
do you mean stages or branches
Answer:
110 N
Explanation:
When a force is applied on a body and body does not move, it means the body remains at rest.
In this condition, there is a contact force between the body and the floor which is called static friction.
Th static friction force is a self adjusting force and comes into play when the body is at rest.
Here, the applied force is 110 N and the chest is not moving, that means a static friction force is acting between the chest and the floor. This static friction force is the force of contact between the chest and the floor. The static friction force is equal to the applied force when the body does not move.
So, the contact force between the chest and the floor is 100 N.
