Difference between revisions of "Net Force"

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Main Page > Giancoli Physics (5th ed) Solutions > Blank Problem Template

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Problems

1. What is the weight of a 3.4 kg mass?

F = ma

Weight = force

F = 3.4 x 9.8 = 33.32 N

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2. What mass has a weight of 720. N?

F = ma

F/a = m

720 N/9.81 m/s = 73.4 m/s

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3. Bob must exert 240. N of force on a 980. Kg car to move it at a constant speed up an incline. (The frictional and gravity force is 240. N) What is the acceleration of the car if he exerts a force of 350. N?

F = ma

F/m = a

(350 N - 240 N)/980 kg = .112 m/s²

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4. What is the acceleration of a 6.0 Kg object hanging on a string that is under a tension of 80. N? 30. N? (Make up positive)

F = ma

6 kg * 9.81 m/s = 58.9 N

A. (80 N - 58.9 N) = (6 kg)a

21.1 N = (6 kg)a

a = 3.5 m/s²

B. (30 N - 58.9 N) = (6 kg)a

(-28.9 N) = (6 kg)a

a = -4.8 m/s²

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5. . What force is needed to accelerate a 60.0 Kg cart and rider from rest to 4.20 m/s in 2.50 seconds when the friction force is 24.0 N?

v/t = a

(4.2 m/s)/(2.5 s) = 1.68 m/s²

F = ma

(F - 24 N) = (60 kg)(1.68 m/s²)

F = 125 N

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6. What is the mass of a box that moves at a constant velocity along a surface with a force of 15 N, and accelerates at +4.2 m/s/s when you exert +26 N? (The box is moving in the same direction as the forces)

F = ma

(26 N - 15 N) = m(4.2 m/s²)

m = 2.6 kg

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7. If you exert a force of 60. N on a car, it moves at a constant velocity. (i.e. there is a frictional force of 60. N) What is its mass if when you exert 80. N on it, it accelerates from rest to 2.0 m/s in 100. seconds?

F = ma

(2.0 m/s) / (100 s) = .02 m/s²

(80 kg - 60 kg) = m(.02 m/s²)

m = 1000 kg

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8. A 60.0 Kg rocket accelerates upward from rest reaching a height of 23.4 m in 3.00 seconds. What must be the thrust of the engine?

SUVAT!!!

s = ut + .5at²

23.4 = .5a(3)²

a = 5.2 m/s²

F = ma

(F - (9.81m/s² * 60 kg)) = 60 kg * 5.2 m/s²

F = 901 N

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9. Question

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10. Question

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11. Question

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12. Question

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13. Question

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14. Question

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15. Question

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