Question
In the first 3.8 seconds of a bobsled run, the 57 kg sled is pushed until it reaches a speed of 1.1 m/s. The flat surface has a coefficient of friction of 0.06. What is the acceleration of the sled in this time? What constant force is exerted by the bobsled team?
Answers
Henry
a = V/t = (1.1m/s)/3.8s = 0.2895 m/s^2.
Ws = m * g = 57kg * 9.8N/kg = 558.6 N.=
Weight of the sled.
Fex - Ff = m * a
Fex - 0.06*558.6 = 57*0.2895
Fex - 33.52 = 16.5
Fex = 16.5 + 33.52 = 50.0 N. = Force
exerted.
Ws = m * g = 57kg * 9.8N/kg = 558.6 N.=
Weight of the sled.
Fex - Ff = m * a
Fex - 0.06*558.6 = 57*0.2895
Fex - 33.52 = 16.5
Fex = 16.5 + 33.52 = 50.0 N. = Force
exerted.