A sled 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.