Question

A free body diagram with 5 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline = blank N. The second vector is pointing right, labeled F Subscript f Baseline = 5 N. The third vector is pointing upward, labeled F Subscript N Baseline = blank N. The fourth vector is pointing left, labeled F Subscript p 1 Baseline = 8 N. The fifth vector is pointing up to the left at an angle of 45 degrees, labeled F Subscript p 2 Baseline = 12 N. The upward vector is shorter than the downward vector. The right vector is shorter than the left vector.
Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. The child pushing the sled is exerting a horizontal a force of 8.00 N. There is a force of friction of 5.00 N.

What is the weight of the sled?
N

What is the normal force exerted on the sled? Round the answer to the nearest whole number.
N

What is the acceleration of the sled? Round the answer to the nearest hundredth.
m/s2

Answer 1

Weight: Fg = m g = (30.0 kg)(9.80 m/s^2) = 294 N

Vertical component of the 12.0 N pull: 12.0 cos45° = 12.0(0.7071) = 8.485 N
Normal force (no vertical acceleration): N + 8.485 N − 294 N = 0 → N = 294 − 8.485 = 285.515 ≈ 286 N

Horizontal components:
- pull horizontal = 12.0 cos45° = 8.485 N (left)
- push = 8.00 N (left)
- friction = 5.00 N (right)
Net horizontal force = 8.485 + 8.00 − 5.00 = 11.485 N (left)

Acceleration = Fnet / m = 11.485 N / 30.0 kg = 0.3828 m/s^2 → 0.38 m/s^2 to the left

Answers:
- Weight = 294 N
- Normal force = 286 N
- Acceleration = 0.38 m/s^2 (to the left)