Asled of mass 50 kg is pulled along snow-covered, flat ground. the static friction coefficient is 0.30, and the sliding friction coefficient is 0.10.. a) what does the sled weigh? . b) what force will be needed to start the sled moving? . c) what force is needed to keep the sled moving at a constant velocity? . d) once moving, what total force must be applied to the sled to accelerate it 3.0 m/s/s? . answers: . a. 490 n. b. 150 n. c. 49 n. d. 200 n

A. The weight of the sled is the product of its mass and the acceleration due to gravity, 
                              (50 kg) x (9.81 m/s²) = 490 N
B. The force needed is the product of static friction coefficient and the normal which is equal to the weight,
                                F = (490 N) x 0.30 = 147 N
C. The force is the product of the sliding friction coefficient and the normal force, 
                               F = (490) x (0.10) = 49 N
D. The net force is calculated as,
                           Fnet = ma = F - 49 N
Substituting the values,
                          (50 kg) x (3.0 m/ s²)  = F - 49 N
The value of F is approximately equal to 199 N.