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Net Force Problems

Net Force Problems. There are 2 basic types of net force problems On a horizontal surface, the formula is: F net = F A - F f If it is a “up-down” problem, the formula is: F net = F up - F down F up = applied force F down = weight. “Up – Down”.

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Net Force Problems

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  1. Net Force Problems • There are 2 basic types of net force problems • On a horizontal surface, the formula is: • Fnet = FA - Ff • If it is a “up-down” problem, the formula is: • Fnet = Fup - Fdown • Fup = applied force Fdown = weight

  2. “Up – Down” If a ball is thrown upward with a force of 148 N and has a mass of 10 kg, what will it’s acceleration be? Fnet = Fup - Fdown Fup Fdown

  3. “Up – Down” If a ball is thrown upward with a force of 148 N and has a mass of 10 kg, what will it’s acceleration be? Fnet = Fup - Fdown Fnet = 148 N - (10 kg)(9.8m/s2) Fup Fdown

  4. “Up – Down” If a ball is thrown upward with a force of 148 N and has a mass of 10 kg, what will it’s acceleration be? Fnet = Fup - Fdown Fnet = 148 N - (10 kg)(9.8m/s2) F net = 148 N - 98 N Fup Fdown

  5. “Up – Down” If a ball is thrown upward with a force of 148 N and has a mass of 10 kg, what will it’s acceleration be? Fnet = Fup - Fdown (m)(a) = 148 N - (10 kg)(9.8m/s2) 10 kg(a) = 148 N - 98 N a = +5 m/s2( upward direction) Fup Fdown

  6. “Up – Down” Another type of “up –down” problem is where the acceleration is given and you need to find the “up” force. A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/ss. What force is needed to do this? Fup Fdown

  7. “Up – Down” A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/ss. What force is needed to do this? Fnet = Fup - Fdown Fup Fdown

  8. “Up – Down” A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/ss. What force is needed to do this? Fnet = Fup - Fdown (m)(a) = Fup - (m)(g) Fup Fdown

  9. “Up – Down” A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/ss. What force is needed to do this? Fnet = Fup - Fdown (m)(a) = Fup - Fdown (2.5 kg)(3.5 m/s2) = Fup - (2.5kg)(9.8m/s2) Fup Fdown

  10. “Up – Down” A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/ss. What force is needed to do this? Fnet = Fup - Fdown (m)(a) = Fup - Fdown (2.5 kg)(3.5 m/s2) = Fup - (2.5kg)(9.8m/s2) 8.75 N = Fup - 24.5N Fup Fdown

  11. “Up – Down A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/ss. What force is needed to do this? Fnet = Fup - Fdown (m)(a) = Fup - Fdown (2.5 kg)(3.5 m/s2) = Fup - (2.5kg)(9.8m/s2) 8.75 N = Fup - 24.5N Fup = 33.25 N Fup Fdown

  12. Horizontal Problems • In order to do these, we need to understand the force of friction first. • Friction.ppt

  13. Horizontal Problems • There are two types of horizontal problems • One is where the applied force is also horizontal • The other type is when the applied force is at an angle with the surface

  14. Horizontal Problems Ff FA 1. FA 2. Ff θ

  15. Horizontal Problems Ff FA 1. All of the applied force is used in the net Force equation 2. FA Ff θ Only the horizontal component of the force is used

  16. Horizontal Problems Ff FA 1. All of the applied force is used in the net Force equation FA 2. Ff θ Fh Only the horizontal component of the force is used

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