John is pushing a wheelbarrow filled with sand on a building site the total mass of the wheelbarrow and its content is 30kg. When he applies force of 20 the wheel barrow moves forward at constant speed. What is the Magnitude of the friction between the wheels and the ground?

Let the magnitude of friction force = f

acceleration ,a = 0 (because it moves with constant speed)
F=ma
20-f=30x0
f=20N

Yes the magnitude will be the same until the force exceed the maximum frictional force then the system will only move with an acceleration.

so what about the gravity because they say between the wheels and the ground and we all know that the ground is the gravity which is 9.8 so dont we include force of gravity when finding the magnitude ?

To find the magnitude of the friction between the wheels and the ground, we need to consider Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma).

In this case, the wheelbarrow is moving at a constant speed, which means its acceleration is zero. Therefore, the net force acting on the wheelbarrow is also zero.

The net force is given by:
Net force = Applied force - Frictional force

Since the wheelbarrow is moving at a constant speed, the applied force and frictional force are equal in magnitude but opposite in direction. So, we can write:

Applied force - Frictional force = 0

Now, plugging in the values given in the problem:
Applied force = 20 N

We can rearrange the equation to solve for the frictional force:
Frictional force = Applied force - 0 (since net force is zero)
Frictional force = 20 N

Therefore, the magnitude of the friction between the wheels and the ground is 20 N.

And curiously, that is the same magnitude of the friction between his shoes and the ground. Think on that, Newton.

Well, John must be one strong guy if he can push a wheelbarrow filled with sand! But let's get to the question at hand. Since the wheelbarrow is moving at a constant speed, we know that the force of friction between the wheels and the ground must be equal in magnitude to the force John applies. In this case, John is applying a force of 20 N, so the magnitude of the friction between the wheels and the ground is also 20 N.