Question

A 15-kg sled is pulled across a horizontal, snow-covered surface by a force applied to a rope at 30 degrees with the horizontal. The coefficient of kinetic friction between the sled and the snow is 0.20. (a) If the force is 33 N, what is the horizontal acceleration of the sled? (b) What must the force be in order to pull the sled at constant velocity?

(a) 2.0 m/s², (b) 33 N
(a) 1.8 m/s², (b) 25 N
(a) 2.5 m/s², (b) 40 N
(a) 1.5 m/s², (b) 30 N

Answer 1

To find the sled's acceleration, calculate the horizontal force component and subtract the friction force. For constant velocity, the horizontal force should equal the friction force. Use trigonometry and Newton's second law for calculations.

Step-by-step explanation:

Calculating the Acceleration and Required Force

For part (a), we calculate the horizontal component of the pulling force using trigonometry. Given the total force of 33 N at a 30-degree angle, the horizontal component (Fhorizontal) is 33 N × cos(30°) = 28.6 N approximately. The frictional force (Ffriction) opposing the motion is found by multiplying the coefficient of kinetic friction by the normal force, which is equal to the weight of the sled since there is no vertical acceleration: Ffriction = μk × m × g = 0.20 × 15 kg × 9.8 m/s² = 29.4 N. The net force is Fnet = Fhorizontal - Ffriction, which we use in Newton's second law, F = ma, to find the acceleration a.

For part (b), the force needed to maintain constant velocity is such that the horizontal component of the pulling force equals the frictional force. Since Ffriction = 29.4 N, we need an equal horizontal force to move the sled at constant velocity. We calculate the total force by dividing the horizontal force by cos(30°).