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?
Science - Henry, Thursday, November 10, 2011 at 5:50pm
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.