Question

A trick or treater applies 3 N net force to drag his 30 kg bag of candy East towards his home, ignoring friction, diagram the forces acting on the bag and solve for the acceleration of the bag.
Draw the free body diagram Include all labels and numerical values. (Recall: you must Calculate Fw)
What is the net force for the X axis?
What is the acceleration?

…

Answer 1

Free-body diagram (text description)

- Draw the bag as a dot or box. Put coordinate axes with +x to the right (East) and +y upward.
- On the dot/box draw:
- Applied force: F_applied = 3 N → (to the right / East)
- Weight: F_w = mg = (30 kg)(9.8 m/s²) = 294 N ↓
- Normal force: F_N = 294 N ↑ (since no vertical acceleration)
- No friction (ignored).

Calculations

1. Weight:
F_w = m g = 30 kg × 9.8 m/s² = 294 N (downward).

2. Net force on x-axis:
ΣF_x = F_applied = 3 N (East).

3. Net force on y-axis:
ΣF_y = F_N − F_w = 0 ⇒ F_N = 294 N.

4. Acceleration (x-direction):
a = ΣF_x / m = 3 N / 30 kg = 0.10 m/s² → East.

Answers
- Net force for the x-axis: 3 N (East).
- Acceleration: 0.10 m/s² (East).