Asked by ww

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is 2.4 N. The free-body diagram shows the forces acting on the sled.

A free body diagram with 4 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline. The second vector is pointing up to the right at an angle of 50 degrees, labeled F Subscript p Baseline. The third vector is pointing upward, labeled F Subscript N Baseline. The fourth vector is pointing left, labeled F Subscript f Baseline. The up and down vectors are the same length. The right vector is longer than the left vector.

What is the acceleration of the sled and the normal force acting on it, to the nearest tenth?

a = 1.3 m/s2; FN = 63.1 N
a = 1.6 m/s2; FN = 65.6 N
a = 1.9 m/s2; FN = 93.7 N
a = 2.2 m/s2; FN = 78.4 N

All Answers 1

Answered by GPT-5 mini AI
Horizontal: Fx = 20 cos50° − 2.4 = 20(0.6428) − 2.4 = 12.8556 − 2.4 = 10.4556 N
a = Fx / m = 10.4556 / 8 = 1.30696 ≈ 1.3 m/s²

Vertical (no vertical acceleration): N + 20 sin50° − mg = 0 ⇒ N = mg − 20 sin50°
mg = 8(9.8) = 78.4 N; 20 sin50° = 20(0.7660) = 15.3208 N
N = 78.4 − 15.3208 = 63.0792 ≈ 63.1 N

Answer: a = 1.3 m/s² and FN = 63.1 N (first choice).