The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Newtons. What force must be exerted to accelerate the box at 0.5 meters per second to the south?

We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

[tex]f=N*miu[/tex]

where:

N = normal force [N]

miu = friction coefficient

f = friction force = 22 [N]

Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

[tex]F - f = m*a[/tex]

where:

F = force exerted [N]

f = friction force [N]

m = mass = 95 [kg]

a = acceleration = 0.5 [m/s²]

Now replacing:

[tex]F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N][/tex]

The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Newtons. What force must be exerted to accelerate the box at 0.5 meters per second to the south?​

Respuesta :

Otras preguntas

The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Newtons. What force must be exerted to accelerate the box at 0.5 meters per second to the south?