AP Physics Questions...

Test Preparation
1

<p>I have been stuck on these questions for quite some time. I know they will require the force formula where F=ma, but what other formula(s) are required to do these, and how should I approach each problem?</p>

<li><p>What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?</p></li>
<li><p>What average force is needed to accelerate a 7.00-gram pellet from rest to 175 m/s over a distance of 0.700 m along the barrel of a rifle?</p></li>
<li><p>A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s^2 using very light fishing line that has a “test” value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?</p></li>
<li><p>A 0.140-kg baseball traveling 45.0 m/s strikes the catcher’s mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove?</p></li>
</ol>

<p>Any help to any of these will be appreciated.</p>

2

<p>7) vfinal = vinital - at
0 = 90 km/hr * 1000m/1km <em>1hr/3600 seconds - a</em>8 secs
a = 3.125 m/s^2
F = ma
F = 1100 kg * 3.125 m/s^2
F = 3,437.5 N</p>

3

<p>give the rest a try</p>

4

<p>why is v(final) = 0?</p>

5

<p>becuase its not moving when u stop it at the end</p>

6

<p>For number 8, I tried using that same formula</p>

<p>vfinal = vinital - at
175m/s=0-a(t)</p>

<p>how do i find the time?</p>

7

<p>or should i use a different formula to find the acceleration?</p>

<p>vfinal^2=vinitial^2 + 2ad?</p>

8

<p>I tried doing number 8 on my own.</p>

<p>vfinal^2=vinitial^2 + 2ad
175m/s=0^2 + 2a(0.700m)
a= 125m/s
F=ma
F=(7.00g)(125 m/s)
F=875N</p>

<p>Is this correct>?</p>

9

<p>You should really post this on physicsforums.com under k-12 homework help.</p>

10

<p>almost!!!</p>

<p>have to convert 7 g * 1 kg/1000 g since a Newton is a kg-m/s^2</p>

11

<p>No wonder. My answer seemed very large.</p>

12

<p>What formula do you suggest for number 9?</p>

13

<p>F = ma just plug in if the Force of breaking the line is 22 N and the acceleration is 4.5 m/s^2 then plug in 22N = m * 4.5 to get the MINIMUM weight of the fish that would break the line.</p>

14

<p>That's what I started doing, and I got 4.9 kg. But the back of the textbook has answers for the odd problems only, and it says m>1.5 kg. How can that be?</p>

15

<p>Whoops - I see - Tension!!!</p>

<p>If the Tension is acting upward, the force of the weight of the fish is acting downards, so the formula is:</p>

<p>Ftension - (mg)of fish = ma the acceleration is only due to the net Force - you have to overcome the weight of the fish before you can accelerate it out of the water.</p>

<p>22 - m(9.8) = m(4.5)
22 = m(14.3)
m minimum of fish = 1.538 kgs</p>

<p>Sorry about that - I better get to sleep...</p>

16

<p>
[quote]
Whoops - I see - Tension!!!</p>

<p>If the Tension is acting upward, the force of the weight of the fish is acting downards, so the formula is:</p>

<p>Ftension - (mg)of fish = ma the acceleration is only due to the net Force - you have to overcome the weight of the fish before you can accelerate it out of the water.</p>

<p>22 - m(9.8) = m(4.5)
22 = m(14.3)
m minimum of fish = 1.538 kgs

[/quote]
</p>

<p>that is correct..</p>

<p>
[quote]
You should really post this on physicsforums.com under k-12 homework help.

[/quote]
</p>

<p>physicsforums.com is a great site for such homework help .. Also, chances are that these questions have already been asked..</p>

<p>For Question 10:</p>

<p>use the formula V^2= v^02 + 2a (d-d0) and take it from there.</p>