posted by Sarah .
A sled mass of 40 kg is pulled along a snow covered surface. The coefficient of static friction is 0.40 and the coefficient of sliding friction is 0.25
a) what force is needed to start the sled moving?
b)what force is needed to start to keep the sled moving at a constant speed?
c) once moving, what total force must be applied to the sled to accelerate it at 3.0 m/s2
M*g = 40*9.8 = 392 N. = Wt. of sled.
Fp = 392*sin 0 = 0 = Force parallel to the surface.
Fn = 392*Cos 0 = 392 N. = Force perpendicular to the surface.
Fs = u*Fn = 0.4 * 392 = 156.8 N. = Force
of static friction.
Fk = 0.25 * 392 N. = 98 N. = Force of kinetic friction.
a. Fap-Fs = M*a
Fap-156.8 = M*0
Fap = 156.8 N. = Force applied.
b. Fap-Fk = M*a
Fap-98 = M*0
Fap = 98 N.
c. Fap-Fk = M*a
Fap-98 = 40 * 3
Fap = 120 + 98 = 218 N.