A 10 kg block is pushed across a rough surface by a 25N force the block moves at a constant velocity when this force is applied.

A) determine the magnitude of the friction force acting on the block
B) What is the coefficient of friction between the surface and the block

Fb = 10kg * 9.8N/kg = 98N.

Fb =98N @ 0 deg.

Fp = 98sin(0) = 0 = Force parallel to plane(hor).

Fv=98cos(0) = 98N.=Force perpendicular to the plane.

a. Fn = Fap - Fp - Ff = 0,
25 - 0 - Ff = 0,
Ff = 25N = Force of friction.

b. Ff = uFv = 25,
u*98 = 25,
u = 0.26 = Coefficient of friction.

To determine the magnitude of the friction force acting on the block, we need to use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration. Since the block is moving at a constant velocity, we know that the acceleration is zero. Therefore, the net force acting on the block is also zero.

The net force acting on the block can be calculated by subtracting the force of friction from the applied force (25N). So, we have:

25N - Force of friction = 0

Therefore, the force of friction acting on the block is 25N.

Now, let's move on to determining the coefficient of friction between the surface and the block. The coefficient of friction (μ) is a measure of how rough or smooth the surface is and can be calculated using the formula:

Force of friction = μ × Normal force

The normal force is the force exerted by a surface to support the weight of the object resting on it. In this case, the normal force is equal in magnitude but opposite in direction to the force of gravity acting on the block, which can be calculated as:

Normal force = mass × gravity

Normal force = 10 kg × 9.8 m/s^2 = 98 N

Now, substituting the values into the formula for the force of friction, we have:

25N = μ × 98N

To find the coefficient of friction (μ), we rearrange the equation:

μ = (25N) / (98N)

μ ≈ 0.255

So, the coefficient of friction between the surface and the block is approximately 0.255.