A crate of mass 180 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, and the coefficient of kinetic friction is 0.36. You push the crate with a horizontal force of 300 N. What is the magnitude of the frictional force on the crate by the floor in this situation

Frictional force on the crate by the floor = 300 N

Explanation:

Let's start off by checking how much frictional force acts on the crate  by the floor both when it is static, and when it is moving.

Weight of the crate: 180 x 9.81 = 1765.8 N

Lets use the equation:  

For static friction:

F = 0.4 x 1765.8

F = 706.32 N

For kinetic friction:

F = 0.36 x 1765.8

F = 635.69 N

Since the force we apply (300 N) is less than the static friction on the crate,

the crate will not move. When the crate does not move, the resultant forces acting on it are 0 N. This means that friction is opposing our 300 N force with 300 N frictional force. Thus friction force is 300 N in this situation.

pe = weightxheight = 2000x1000=2,000,000joules

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is