
Bob M.
asked 11/10/19Fnet=Fma problems
1) a 500 kg car is travelling at 15 m/s and experiences a forward engine force of 2000N for 3 seconds. The combined frictional forces acting on the car are 600 N. What is the car's acceleration?
2) Four people are pushing on a box where the box has no friction but their feet do. Person #1 pushes with 22N to the right, person #2 with 14 N to the right, and and person #3 pushes with 30 N to the right. Person #4 pushes with 72 N to the left. The box has a mass of 5 kg; determine acceleration.
3) Edward applied a horizontal force to stop a pickup truck that was sliding toward Bella. Find the magnitude of the horizontal force needed, if the truck, going from an initial speed of 14 m/s, stopped over a distance of 1.0 m (the length of Edward's arms). The truck's mass is 1600 kg.
1 Expert Answer
Strategy: Notice that these are all motion-in-one-dimension problems. As with all problems in physics, you can apply Six Steps to Problem Solving, as follows:
- Draw a diagram or graph. Learning how to draw an accurate free-body diagram can help you identify the various forces acting on the object.
- Identify your goal. What is the question asking?
- Identify the data. What do you know? What do you need?
- Develop a math expression (or set of expressions) that captures everything you need. Also, estimate the range of your solution.
- Solve it.
- Check it. Does it match your estimate? Are the units accurate? Did you answer the question/s?
Let's investigate the first question, as an example.
Draw a diagram
<--- 600 Newtons -- [object] -- 2,000 Newtons -------------->
Identify the goal
What is the car's acceleration?
Identify your data
Force due to friction = -600 N
Applied force = +2,000 N
mass = 500 kg
Initial velocity doesn't matter for this question, since you are only trying to find the acceleration. You should first identify all the forces, then add them together. Once you have the vector sum of all forces, then you can use this net force (your resultant) and the mass to find the acceleration.
Set up a math expression
Fnet = Ffriction + Fapplied
Fnet = m*a therefore a = Fnet
m
Solve
Fnet = -600 N + 2,000 N
= 1,400 N in the positive direction
and
a = Fnet / m
= 1,400 N / 500 kg
Note that this means the units are kg*m / s2 divided by kg
a = 2.8 m/s2
For your problems 2 and 3, notice that you will need to follow a similar problem-solving method. First, draw a diagram. Then, identify your question. Third, identify your data. Fourth, reason through the problem. Write a set of math expressions that captures all the data. AFTER you have identified the relationships described in this set of math expressions, THEN you solve by plugging in values. Don't plug values until you have already reasoned through the problem! After you have solved the set, then re-read the problem statement to make sure you have answered the question. Carefully check each part of the problem and check your units to make sure you have verified accuracy and eliminated sources of potential error. Then check the value to make sure it makes sense. Finally, express your solution with the proper units and provide any necessary explanation.
With problems such as these, you will always be provided the data you need; and you may be provided with additional data that is not necessary. You are expected to figure out what you need and select from those. In the first problem, there were several potential distractors: the time over which the force is applied; the initial velocity of the object. These data could help you find things like momentum and final velocity, but you are not asked to identify those.
Let me know if you have any further questions about this problem set, and I'll be glad to help. It's pretty exciting to find out that motion can be described neatly in math language! I hope you'll enjoy solving these.
Best wishes,
Lindsey
Bob M.
Thank you for responding. I was able to solve for the first two questions, but I'm a bit confused on the third. What does it mean by magnitude?11/14/19

Lindsey K.
09/01/23
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Mark H.
I don't understand "Fnet = Fma" We have a common expression "F = ma", where F is sometimes the "net force" resulting from several different conditions.11/10/19