An Olympic bobsledder pushes a 200 kg bobsled at 8 m/s. If the man pushes with a force of 1200 N, how much force of friction is present?

Wb = mg = 200kg * 9.8N/kg = 1960N.

Fb = 1960N @ 0deg. = Force of bobsled.
Fp = 1960sin(0) = 0 = Force parallel to surface.
Fv = 1960cos(0) = 1960N. = Force perpendicular to surface.

Fn = Fap - Fp - Ff = ma = 0, a = 0.
1200 - 0 - Ff = 0,
Ff = 1200 - 0 = 1200N. = Force of friction.

To determine the force of friction, we can use 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. We can find the acceleration by dividing the net force by the mass.

Given data:
Mass of the bobsled (m): 200 kg
Speed of the bobsled (v): 8 m/s
Force applied (F): 1200 N

First, let's calculate the acceleration of the bobsled using Newton's second law:

Acceleration (a) = Net force (F) / Mass (m)

a = F / m
= 1200 N / 200 kg
= 6 m/s²

Now, we need to determine the force of friction. The force of friction can be calculated using the equation:

Force of friction (Ffriction) = mass (m) × acceleration due to friction (afriction)

Since the bobsled is moving uniformly, the acceleration due to friction is equal to the acceleration of the bobsled (a). Therefore:

Ffriction = m × a
= 200 kg × 6 m/s²
= 1200 N

Hence, the force of friction acting on the bobsled is 1200 N.