a 12 kg sled is pulled horizontally with a force of 82 N. the coefficent of the kentic friction of the sled is 0.42. draw a FBD of the sled.
2 answers
how do u find the acceleration of the sled? and what applied force would be required to cause the sled to move at a constant velocity, explain your logic and show all your calculations.
Fs = mg = 12kg * 9.8N/kg = 117.6N = Weight of sled.
Fp = 117.6sin(o) = o. = Force parallel
to plane due to sled.
Fv = 117.6cos(o) = 117.6N. = Force
perpendicular to plane due to sled.
Ff = u*Fv = 0.42 * 117.6 = 49.4N =
Force due to Kinetic friction.
Fn = Fap - Ff = ma,
82 - 49.4 = 12a,
12a = 32.61,
a = 2.72m/s^2.
Fp = 117.6sin(o) = o. = Force parallel
to plane due to sled.
Fv = 117.6cos(o) = 117.6N. = Force
perpendicular to plane due to sled.
Ff = u*Fv = 0.42 * 117.6 = 49.4N =
Force due to Kinetic friction.
Fn = Fap - Ff = ma,
82 - 49.4 = 12a,
12a = 32.61,
a = 2.72m/s^2.
3 answers