Solutions to Net Force

Well, if it moves at a constant velocity when you exert a force of 15 N on it, then it is not accelerating, and the net force is zero on the object. That means that your applied force is counteracted by another force, perhaps friction of 15 N. First set up Newton's second law:
F = ma <26 N- 15 N> = m (4.2 m/s/s)
m = 11/4.2
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?

First find the acceleration:
a_{av =}Dv/Dt     a_av = 2.0/100. = .020 m/s/s
If it moves at a constant velocity when you exert a force of 60. N on it, then it is not accelerating, and the net force is zero on the object. That means that your applied force is counteracted by another force, perhaps friction of 60 N.
Next set up Newton's second law
F = ma
<80 N - 60 N> = m(.020 m/s/s)
m = 1000 N

8. A 60 Kg rocket accelerates upward from rest reaching a height of 23.4 m in 3.0 seconds. What must be the thrust of the engine?

First find the acceleration:
x = 23.4 m
vi = 0
vf = don't care
a = ?????
t = 3.0 s
Use:
x = v_it + ¹/₂at² 23.4 = (0 x 3) + .5 a 3.0²     23.4 = 4.5a   a = 5.2 m/s/s
Next find the weight:
weight = 60 x 9.8 = 588
And finally, set up Newton's second law:
<F - 588N ) = (60. kg)(5.2 m/s/s)
F - 588 = 312
F = 900 N
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9. It takes 45 N to make a 10. kg cart move at a constant speed. What force does it take to make the cart accelerate at 3.2 m/s/s in the direction it is moving?

If it moves at a constant velocity when you exert a force of 45. N on it, then it is not accelerating, and the net force is zero on the object. That means that your applied force is counteracted by another force, perhaps friction of 45 N.
So set up Newton's second law:
F = ma
<F - 45 N> = (10. kg)(3.2 m/s/s)
F - 45 = 32
F = 77 N

10. What tension would accelerate a 5.0 Kg object suspended on a string upwards at 6.2 m/s/s? Downwards?

Find the weight first:
weight = 5.0 x 9.8 = 49 N
Next, set up Newton's second law. When we are accelerating up, the acceleration is positive:
F = ma
<F - 49 N> = (5.0 kg)(+6.2 m/s/s)
F - 49 = 31
F = 80 N
When we are accelerating down, the acceleration is negative:
F = ma
<F - 49 N> = ( 5.0 kg)(-6.2 m/s/s)
F - 49 = -31
F =18 N

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11. A 45.0 gram rocket accelerates upward from 0 to 12.0 m/s in .0500 seconds What must be the thrust of the engines?

mass = 45 g = .045 kg (divide by 1000)
First find the acceleration:
a_{av =}Dv/Dt
a_av = 12/.05 = 240 m/s/s
Next find the weight:
weight = .045 x 9.8 = .441 N
Finally, set up Newton's second law:
F = ma
<F - .441 N> = (.045 kg)(240 m/s/s)
F = 11.241 N = 11.2 N

12. A rocket has engines that produce 60.0 N of thrust. What is its mass if it accelerates upward at 40.0 m/s/s?

First, find the weight:
F = ma
weight = m(9.80 m/s/s) - don't worry that you don't know the mass, just leave it as a variable.
Next, set up Newton's second law:
<60.0 - 9.80m> = m(40.0)
60 = 9.80m + 40.0m = 49.8m
m = 1.2 kg

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13. A dog is pulling forward on a 215 kg sled that slows from +6.20 m/s to rest in a distance of 8.25 m. What is the deceleration of the sled? If the force of friction slowing the sled is 782 N, what force is the dog exerting in the direction the sled moves?

Let's solve the kinematics problem for the acceleration:

x = 8.25 m
vi = 6.20 m/s
vf = 0
a = ??
t = don't really care too much

use v_f² = v_i² + 2ax:
(0)² = (6.20 m/s)² + 2a(8.25 m)

a = -2.3297 m/s/s = -2.33 m/s/s

Now we are ready to solve Newton's second law. We have the dogs exerting an unknown force (presumably to the right) of F, the force of friction of 782 N acting to the left (-):
<F> = ma
<F - 782 N> = (215 kg)(-2.3297 m/s/s)
F = 281.11515 N = 281 N

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14. A 18,380 kg airplane slows from 48.1 m/s to rest in 6.14 seconds. What was its acceleration? If the engines generated 112 kN (112,000 N) of reverse thrust, how much air friction was acting against the plane as it slowed down? (this would be average)

Let's solve the kinematics problem for the acceleration:

x = don't give a hoot
vi = 0 m/s
vf = 48.1 m/s
a = ??
t = 6.14 s

Using v_f = v_i + at:
48.1 m/s = 0 + a(6.14 s)
a = -7.83388 m/s/s = -7.83 m/s/s

Now Newton's second law has the reverse thrust acting to the left as the plane moves to the right - in the positive direction, and some unknown force of friction (F) acting presumably in the negative direction as well. I will put it in here as positive and let the math tell me whether it is really negative:
<F> = ma
<-112,000 N + F> = (18,380 kg)(7.83388 m/s/s)
F = -31986.64495 N = -32.0 kN

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15. A drop tower has a 118 kg experiment that free falls from rest for 2.20 seconds, and strikes an airbag that slows it to rest in a distance of 3.20 m. With what velocity does the experiment strike the airbag? What is the upward acceleration as the experiment stops? What is the upward force acting on the experiment to stop it?

Let's solve the kinematics problem for the velocity of impact:

x = don't give a hoot
vi = 0 m/s
vf = ???
a = -9.80 m/s/s
t = 2.20 s

Using v_f = v_i + at:
vf = 0 + (-9.80 m/s/s)(6.14 s)
v = -21.56 m/s = -21.6 m/s

Now let's solve for the deceleration that the airbag causes in stopping the experiment. The experiment goes from an initial velocity of -21.56 m/s to rest in a downward displacement of -3.20 m:

x = -3.20 m
vi = -21.56 m/s
vf = 0
a = ???
t = I am not caring

use v_f² = v_i² + 2ax:
(0)² = (-21.56 m/s)² + 2a(-3.20 m)
a = +72.63025 m/s/s = +72.6 m/s/s. (it is accelerating upward as it stops going downward)

Now we can solve Newton's second law for the experiment as it stops. We have the force of the airbag (F) acting up on the experiment, and its weight of (118 kg)(9.80 N/kg) = 1156.4 N acting downward:
<F> = ma
<F - 1156.4 N> = (118 kg)(+72.63025 m/s/s)
F = 9726.7695 = 9730 N

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Solutions to Net Force

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