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 (labels and values)

- Draw the bag as a dot/box. From that dot:
- To the right (East): Applied force F_ap = 3.0 N
- Downward: Weight F_w = mg = 30 kg × 9.8 m/s^2 = 294 N
- Upward: Normal force N = 294 N (since no vertical acceleration)
- (No friction)

Equations and numbers
- Vertical: ΣF_y = N − mg = 0 ⇒ N = mg = 30(9.8) = 294 N
- Horizontal: ΣF_x = F_ap = 3.0 N (East)

Net force on X axis
- ΣF_x = +3.0 N (positive = East)

Acceleration
- a = ΣF_x / m = 3.0 N / 30 kg = 0.10 m/s^2 East

Answers
- Net force (x): 3.0 N East
- Acceleration: 0.10 m/s^2 East