F=ma Practice Problems

F=ma Problem Set

Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations.

In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems.

F=ma Equations

1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?

1

a = 33.33 m/s2 Right

2. What is the acceleration of the 25 kg box that has 50 N of force applied to the right?

2

               a=2.0 m/s2 Right

3. What is the acceleration of the 3 kg box that has 25 N of force applied to the right and 55 N left?

3

              a = 10.0 m/s2 Left

4. What is the acceleration of the 5 kg box that has a 25 N force and 50 N force applied both right?

4

            a = 15.0 m/s2 Right

5. What is the acceleration of the 25 kg box that has a 100 N force north and 50 N force east applied?

5

             a= 4.47 m/s2

5b. What direction would this box accelerate?

            63.43° North of East

6. Does a 795 kg Lorinser speedy, 6300 kg elephant, or 8.6 kg wagon have more inertia and why?

            6300 kg Elephant

            The more mass the more inertia

different masses

7. How much force is required to accelerate a 795 kg Lorinser Speedy by 15 m/s2?

            F = 11925 N

8. How much force is required to accelerate an 8.6 kg wagon by 15 m/s2?

            F = 129 N

9. How much does a 6300 kg elephant accelerate when you apply 500 N of force?

            a = 0.0794 m/s2

10. What is the mass of an object if it takes a net force of 40 N to accelerate at a rate of 0.88 m/s2?

            m = 45.45 kg

11. How much force is required to accelerate a 0.142 kg baseball to 44.7 m/s during a pitchers 1.5 meter delivery?

            F = 94.58 N

12. A 0.050 kg golf ball leaves the tee at a speed of 75.0 m/s. The club is in contact with the ball for 0.020 s. What is the net force of the club on the ball?

               F = 187.5 N

13. A 90.0 kg astronaut receives a 30.0 N force from her jetpack. How much faster is she be moving after 2.00 seconds?

            0.667 m/s faster

14. A 795 kg car starts from rest and travels 41 m in 3.0 s. How much force did the car engine provide?

            F = 7242 N

15. Joe and his sailboat have a combined weight of 450 kg. How far has Joe sailed when he started at 5 m/s and a gust of wind provided 600 Newtons of force for 4 seconds?

            x = 30.64 m

16. Tom pulls a 45 kilogram wagon with a force of 200 Newtons at a 15° angle to the horizontal from rest. How much faster will the wagon be moving after 2 seconds?

force at an angle

            vf = 8.58 m/s

Links

Terms and Conditions - Privacy Policy