assts 10

course Phy 121

wWՎȇҀұvWx{VZassignment #009

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

009. `query 9

Physics I

06-27-2007

......!!!!!!!!...................................

12:32:34

Introductory prob set 3 #'s 1-6 If we know the distance an object is pushed and the work done by the pushing force how do we find the force exerted by the object?

......!!!!!!!!...................................

RESPONSE -->

force = distance * work

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:33:33

If we know the net force exerted on an object and the distance through which the force acts how do we find the KE change of the object?

......!!!!!!!!...................................

RESPONSE -->

KE=F * 'ds

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:35:13

Why is KE change equal to the product of net force and distance?

......!!!!!!!!...................................

RESPONSE -->

KE is the result of the change that the F does for so long a distance

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:36:12

When I push an object with a constant force, why is KE change not equal to the product of the force I exert and the distance?

......!!!!!!!!...................................

RESPONSE -->

because the object had an unknown KE to begin with

confidence assessment: 1

.................................................

veFֿi

assignment #010

010. `query 10

Physics I

06-27-2007

......!!!!!!!!...................................

20:56:06

Query introductory problem set 3 #'s 7-12

Describe two ways to find the KE gain of an object of known mass under the influence of a known force acting for a given time, one way based on finding the distance the object moves and the other on the change in the velocity of the object, and explain why both approaches reach the same conclusion.

......!!!!!!!!...................................

RESPONSE -->

you have to find out the velocity first, you can do that knowing either the 'ds or the 'dv as long as you have the mass and the 'dt. then the KE is equal to .5m * v^2

confidence assessment: 3

.................................................

......!!!!!!!!...................................

20:59:41

General College Physics and Principles of Physics: prob 2.04 convert 35 mi/hr to km/hr, m/s and ft/s.

......!!!!!!!!...................................

RESPONSE -->

35m/hr = 56.315km/hr (35 * 1.609)

56315m/hr = 56315m/3600s= 15.64m/s

=184800ft/hr = 184800ft/3600s = 51.33 ft/s

confidence assessment: 3

.................................................

......!!!!!!!!...................................

21:01:53

Gen phy and prin phy prob 2.16: sports car 95 km/h stops in 6.2 s; find acceleration

......!!!!!!!!...................................

RESPONSE -->

v0=95km/hr

vf=0

vAve=47.5 km/hr

aAve='dv/'dt

a=95/6.2

a=15.3km/hr^2

confidence assessment: 3

.................................................

Čawwbn{~

assignment #010

010. Note that there are 10 questions in this set.

.Force and Acceleration.

Physics I

06-27-2007

......!!!!!!!!...................................

13:28:26

`q001. If a block of mass 10 kg accelerates at 2 m/s^2, then what net force is acting on the block?

......!!!!!!!!...................................

RESPONSE -->

10kg * 2= 20N is the net force

confidence assessment: 3

.................................................

......!!!!!!!!...................................

13:29:12

`q002. How much force must be exerted by someone pulling on it to accelerate a 10 kg object at 2 m/s^2?

......!!!!!!!!...................................

RESPONSE -->

20 Newtons

confidence assessment: 3

.................................................

......!!!!!!!!...................................

13:29:54

`q003. If friction exerts a force of 10 Newtons in the direction opposite the motion of a 10 kg object, then how much force must be exerted by someone pulling on it to accelerate the 10 kg object at 2 m/s^2, with the acceleration in the same direction as the motion?

......!!!!!!!!...................................

RESPONSE -->

30 Newtons

confidence assessment: 2

.................................................

......!!!!!!!!...................................

13:30:47

`q004. How can we write an equation to solve this problem? Hint: What equation would relate the net force Fnet, the force F exerted by the person and the force fFrict of friction?

......!!!!!!!!...................................

RESPONSE -->

netF=m*a + f

confidence assessment: 3

.................................................

......!!!!!!!!...................................

13:31:32

`q005. If a constant net force of 12 Newtons acts on a cart of mass 6 kg, then at what rate does the velocity of the cart change?

......!!!!!!!!...................................

RESPONSE -->

12/6= 2m/s

confidence assessment: 2

.................................................

......!!!!!!!!...................................

13:32:28

`q006. If a force of 50 Newtons is exerted in the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object?

......!!!!!!!!...................................

RESPONSE -->

50 - 10=40

40/20=

2m/s^2

confidence assessment: 3

.................................................

......!!!!!!!!...................................

13:33:27

`q007. If a force of 50 Newtons is exerted opposite to the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object?

......!!!!!!!!...................................

RESPONSE -->

would it be -2m/s^2?

confidence assessment: 2

.................................................

......!!!!!!!!...................................

13:37:03

`q008. If a 40 kg object is moving at 20 m/s, then how long will a take a net force of 20 Newtons directed opposite to the motion of the object to bring the object to rest?

......!!!!!!!!...................................

RESPONSE -->

F=m*a

800N=40kg*a

a=20m/s

i am guessing 4 sec

confidence assessment: 1

.................................................

......!!!!!!!!...................................

13:39:13

`q009. If we wish to bring an object with mass 50 kg from velocity 10 m/s to velocity 40 m/s in 5 seconds, what constant net force would be required?

......!!!!!!!!...................................

RESPONSE -->

F=m*a

F=50kg*30m/s

F=1500 N / 5s

=300N

that doesn't sound right

confidence assessment: 0

.................................................

......!!!!!!!!...................................

13:40:41

`q010. If a constant net force of 50 Newtons brings an object to rest in four seconds from an initial velocity of 8 meters/second, then what must be the mass of the object?

......!!!!!!!!...................................

RESPONSE -->

50=m*-2m/s^2

50=25 * 2

mass=25kg

confidence assessment: 2

.................................................

"

Per my previous notes, you'd better run through the program again and be sure you respond to the given solutions so that they appear in your SEND file.

You already know how to do this but for reference:

&#To run the program correctly you click on the Next Question/Solution button (top left), enter your response in the Answer box (the box at top right), click on the Enter Response button (above the top right box), then again on the Next Question/Solution botton (top left) to see the solution, and finally enter your response or self-critique in the Answer box and click on Enter Response.

You then start the process over with the next question, clicking on the Next Question/Solution button at top left, etc.

&#