Difference between revisions of "Net Force"

From TuHSPhysicsWiki
Jump to: navigation, search
(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?)
 
(10 intermediate revisions by one other user not shown)
Line 1: Line 1:
[[Main Page]] > [[Giancoli Physics (5th ed) Solutions]] > Blank Problem Template<br />
+
[[Main Page]] > [[Solutions to Worksheets]] > Net Force<br />
 
<br />
 
<br />
 
==About this page==
 
==About this page==
All images uploaded for this page must start with the string "Gp5_chapter#_" so  the image 16-38.jpg associated with chapter 16 should be uploaded as Gp5_16_16-38.jpg.  This way we can avoid conflicts in the image directory, and we can find images easily.<br>
+
All images uploaded for this page must start with the string "Net_Force_" so  the image 16-38.jpg associated with chapter 16 should be uploaded as Net_Force_16-38.jpg.  This way we can avoid conflicts in the image directory, and we can find images easily.<br>
 
[[#top | Table of Contents]]
 
[[#top | Table of Contents]]
 
----
 
----
 +
 
==Problems==
 
==Problems==
 
===1. What is the weight of a 3.4 kg mass?===
 
===1. What is the weight of a 3.4 kg mass?===
Line 94: Line 95:
 
----
 
----
  
===9. Question===
+
===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?===
 
<blockquote>
 
<blockquote>
Solution goes here
+
F = ma<br><br>
 +
(F - 45 N) = 10 kg * 3.2 m/s<sup>2</sup><br><br>
 +
F = 77 N
 
</blockquote>
 
</blockquote>
 
[[#top | Table of Contents]]
 
[[#top | Table of Contents]]
 
----
 
----
===10. Question===
+
 
 +
===10. What tension would accelerate a 5.0 Kg object suspended on a string upwards at 6.2 m/s/s?  Downwards?===
 
<blockquote>
 
<blockquote>
Solution goes here
+
F = ma<br><br>
 +
'''A.'''  (F - (5 kg * 9.81 m/s<sup>2</sup>) = 5 kg * 6.2 m/s<sup>2</sup><br><br>
 +
F = 80 N<br><br>
 +
'''B.'''  (F - (5 kg * 9.81 m/s<sup>2</sup>) = 5 kg * -6.2 m/s<sup>2</sup><br><br>
 +
F = 18 N
 
</blockquote>
 
</blockquote>
 
[[#top | Table of Contents]]
 
[[#top | Table of Contents]]
 
----
 
----
===11. Question===
+
 
 +
===11. A 45.00 gram rocket accelerates upward from 0 to 12.00 m/s in .05000 seconds.  What must be the thrust of the engines?===
 
<blockquote>
 
<blockquote>
Solution goes here
+
SUVAT!!! AGAIN!!!<br><br>
 +
v = u + at<br><br>
 +
12 m/s = 0 m/s + a(.05 s)<br><br>
 +
a = 240 m/s<sup>2</sup><br><br>
 +
F = ma<br><br>
 +
(F - (.045 kg * 9.81 m/s<sup>2</sup>)) = .045 kg * 240 m/s<sup>2</sup><br><br>
 +
F = 11.24 N
 
</blockquote>
 
</blockquote>
 
[[#top | Table of Contents]]
 
[[#top | Table of Contents]]
 
----
 
----
===12. Question===
+
 
 +
===12. A rocket has engines that produce 60. N of thrust.  What is its mass if it accelerates upward at 40. m/s/s?===
 
<blockquote>
 
<blockquote>
Solution goes here
+
Here comes a whole bunch of Algebra...<br><br>
 +
F = ma<br><br>
 +
(60 N - (mg)) = m(40 m/s<sup>2</sup>)<br><br>
 +
1.5 N/m/s<sup>2</sup> - mg / 40 m/s<sup>2</sup> = m<br><br>
 +
1.5 N/m/s<sup>2</sup> = mg / 40 m/s<sup>2</sup> + m<br><br>
 +
1.5 N/m/s<sup>2</sup> = mg + 40m / 40 m/s<sup>2</sup><br><br>
 +
60 N = 40m + 9.81m<br><br>
 +
60 N = 49.81m <br><br>
 +
m = 1.2 kg
 
</blockquote>
 
</blockquote>
 
[[#top | Table of Contents]]
 
[[#top | Table of Contents]]
 
----
 
----
===13. Question===
+
 
 +
===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?===
 
<blockquote>
 
<blockquote>
Solution goes here
+
-2.33 m/s/s<br><br>
 +
+281 N<br><br>
 
</blockquote>
 
</blockquote>
 
[[#top | Table of Contents]]
 
[[#top | Table of Contents]]
 
----
 
----
===14. Question===
+
 
 +
===14. An 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)===
 
<blockquote>
 
<blockquote>
Solution goes here
+
-7.83 m/s/s<br><br>
 +
-32.0 kN<br><br>
 
</blockquote>
 
</blockquote>
 
[[#top | Table of Contents]]
 
[[#top | Table of Contents]]
 
----
 
----
===15. Question===
+
 
 +
===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?===
 
<blockquote>
 
<blockquote>
Solution goes here
+
-21.6 m/s<br><br>
 +
+72.6 m/s/s<br><br>
 +
+9750 N<br><br>
 
</blockquote>
 
</blockquote>
 
[[#top | Table of Contents]]
 
[[#top | Table of Contents]]
 
----
 
----

Latest revision as of 17:36, 10 February 2009

Main Page > Solutions to Worksheets > Net Force

Contents

About this page

All images uploaded for this page must start with the string "Net_Force_" so the image 16-38.jpg associated with chapter 16 should be uploaded as Net_Force_16-38.jpg. This way we can avoid conflicts in the image directory, and we can find images easily.
Table of Contents


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

Table of Contents


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

F = ma

F/a = m

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

Table of Contents


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

Table of Contents


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

Table of Contents


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

Table of Contents


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

Table of Contents


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

Table of Contents


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

Table of Contents


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?

F = ma

(F - 45 N) = 10 kg * 3.2 m/s2

F = 77 N

Table of Contents


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

F = ma

A. (F - (5 kg * 9.81 m/s2) = 5 kg * 6.2 m/s2

F = 80 N

B. (F - (5 kg * 9.81 m/s2) = 5 kg * -6.2 m/s2

F = 18 N

Table of Contents


11. A 45.00 gram rocket accelerates upward from 0 to 12.00 m/s in .05000 seconds. What must be the thrust of the engines?

SUVAT!!! AGAIN!!!

v = u + at

12 m/s = 0 m/s + a(.05 s)

a = 240 m/s2

F = ma

(F - (.045 kg * 9.81 m/s2)) = .045 kg * 240 m/s2

F = 11.24 N

Table of Contents


12. A rocket has engines that produce 60. N of thrust. What is its mass if it accelerates upward at 40. m/s/s?

Here comes a whole bunch of Algebra...

F = ma

(60 N - (mg)) = m(40 m/s2)

1.5 N/m/s2 - mg / 40 m/s2 = m

1.5 N/m/s2 = mg / 40 m/s2 + m

1.5 N/m/s2 = mg + 40m / 40 m/s2

60 N = 40m + 9.81m

60 N = 49.81m

m = 1.2 kg

Table of Contents


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?

-2.33 m/s/s

+281 N

Table of Contents


14. An 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)

-7.83 m/s/s

-32.0 kN

Table of Contents


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?

-21.6 m/s

+72.6 m/s/s

+9750 N

Table of Contents