Question

A net applied force of 20 Newtons pushes a grocery cart left across the store. It accelerates at 4.0 m/s2. Neglect friction (Ff is 0N).
SHOW YOUR FORMULA AND WORK
What is the mass of the cart?
Draw a free diagram of the cart. Label and provide # with units for each vector. (use applied force found for 1 m/s2)
------------------------------ What net force would be required for the same object to reduce by a rate of 1 m/s2?
--- What is the applied force now?
What is the weight of a cart?
---Is there a normal, what is?

Answer 1

1) Find the mass
Formula: Fnet = m a
Solve: m = Fnet / a = 20 N / (4.0 m/s^2) = 5.0 kg

So the cart's mass is 5.0 kg.

2) Free-body diagram (use the applied force found for the 1 m/s^2 deceleration case)
(Described vectors with magnitudes and directions)
- Applied force, Fa = 5.0 N, to the right (this is the horizontal force that would produce a = −1.0 m/s^2 if the cart was originally moving left)
- Weight, W = m g = 5.0 kg × 9.8 m/s^2 = 49 N, downward
- Normal force, N = 49 N, upward (surface supports the cart)
- Friction, Ff = 0 N (neglected)
- Acceleration vector, a = 1.0 m/s^2 to the right (magnitude 1.0 m/s^2; direction opposite the cart's original motion to reduce its speed)

3) What net force is required to reduce the speed at 1 m/s^2?
Use Fnet = m a with a = −1.0 m/s^2 (opposite the current velocity):
Fnet = 5.0 kg × (−1.0 m/s^2) = −5.0 N
Interpretation: magnitude 5.0 N, direction opposite the motion (to the right if motion is to the left).

4) What is the applied force now?
With no friction, the applied force = net force, so Fa = 5.0 N to the right.

5) What is the weight of the cart?
W = m g = 5.0 kg × 9.8 m/s^2 = 49 N (downward).

6) Is there a normal force, and what is it?
Yes. On a horizontal floor the normal force balances the weight: N = 49 N upward (so vertical forces sum to zero).