An ∃mazing Physics thread (Unit 2)

[e^1]

Been a while since I posted for Physics help since this Motion chapter has been excruciatingly long. I've asked many this question before and it has been approached in different ways every time. I would like to know how people go about solving it in the most succinct but complete way.
Thanks.

A force of 120 N is used to push a 20 kg shopping trolley along
the line of its handle—at 20° down from the horizontal. This is
enough to cause the trolley to travel with constant velocity to
the north along a horizontal path.

a) What is the value of the frictional force acting against the
trolley?
b) How large is the normal force that is supplied by the ground
on which the trolley is pushed?

Crap misread. Just ignore

a)



This is the horizontal component of weight. Since it is travelling at constant velocity, then the resultant force is zero (according to Newton's First Law of motion). That is, there must be 120 N in the opposite direction that opposes the 120 N travelling forward.

Hence:





b) To find the magnitude of the normal force, simply find the vertical component of weight. Both are equal in magnitude because the object is only moving forward in an inclined plane.




Draw a force diagram if that helps.

[Guest]

a)



This is the horizontal component of weight. Since it is travelling at constant velocity, then the resultant force is zero (according to Newton's First Law of motion). That is, there must be 120 N in the opposite direction that opposes the 120 N travelling forward.

Hence:





b) To find the magnitude of the normal force, simply find the vertical component of weight. Both are equal in magnitude because the object is only moving forward in an inclined plane.




Draw a force diagram if that helps.

Sorry, but I'm afraid that's incorrect my friend.

Answers:
(a) 110N to the south.
(b) 240N upwards

Try again, if you will

[psyxwar]

Your answers are off, lol.

Anyways:

Resolve the vectors in the down and north directions.

Force north: 120 cos 20 = 112.76N
Force down: 120 sin 20 = 41.04N

(a) Net force = 0, it's simply 112.76N south
(b) Normal needs to be equal to force downwards in magnitude.

Force downwards = 41.04 + 20(10m/s^2), where 20 is mass and 10 is acceleration due to gravity
= 241.04N

Therefore normal force is 241.04N up

[Homer]

yeap agree with psyxwar

[Guest]

Great way to bring up his 500th post
Seems like he knows his Physics as well as Chem too.

[Guest]

Last question of the chapter, aaaannnd this puts me down.

In a horrific car crash, a car skids 85 m before striking a parked car in the rear with
a velocity of 15 m s−1. The cars become locked together and skid a
further 5.2 m before finally coming to rest. The mass of the first car,
including its occupants, is 1350 kg. The parked car has a mass of
1520 kg.

a) What is the impulse on each car during the collision?
b) What is the average size of the frictional force between road and
car that finally brings them to rest?

I got I=1.1*10^4 Ns, for question (a). I used the I=mv formula to multiply the second car's mass (1520kg) by the final velocity of the system (7.06m/s/s). So the working was: "1520*7.06=1.1*10^4" Can someone verify if this is correct?
I'm not sure how to work out part (b).

[BasicAcid]

Last question of the chapter, aaaannnd this puts me down.

In a horrific car crash, a car skids 85 m before striking a parked car in the rear with
a velocity of 15 m s−1. The cars become locked together and skid a
further 5.2 m before finally coming to rest. The mass of the first car,
including its occupants, is 1350 kg. The parked car has a mass of
1520 kg.

a) What is the impulse on each car during the collision?
b) What is the average size of the frictional force between road and
car that finally brings them to rest?

I got I=1.1*10^4 Ns, for question (a). I used the I=mv formula to multiply the second car's mass (1520kg) by the final velocity of the system (7.06m/s/s). So the working was: "1520*7.06=1.1*10^4" Can someone verify if this is correct?
I'm not sure how to work out part (b).

Be careful, note how it says for each car, so you also have to do that multiplication for the first car as well. I got what you got.

Part b)
u=7.06
v= 0
s= 5.2
a= ?

v^2 = u^2 + 2as
Thus a = -4.79 (note how it's negative as its coming to a stop)

Now F=ma
F=2870*(-4.79)
=-1.4 * 10^4 N

Only take the magnitude as it asks for 'average' size.

[Guest]

Came across a weird question on the 'Area of Study Review' section. I don't understand the relevance of Hz to time or speed.

The following information applies to questions 1–2. During a physics experiment a student sets a multi-flash timer at a frequency of 10 Hz. a nickel marble is rolled across a horizontal table. The first four flashes are in the order: A,B,C and D. The distance from A to B is 1.0cm. The distance from B to C is 3.0cm. The distance from C to D is 5.0cm.
Assume that when flash a occurred t = 0, at which time the marble was at rest.

1) Determine the average speed of the marble as it traveled from A to B.
2) Determine the instantaneous speeds of the marble when t=0.05 s.

For Question 1, I designated the following values: u=o, x=0.01, t=1/6, v=?
Notice t=1/6. I'm not entirely sure if it should be 1/6. I simply got that by dividing 10 by 60. Is this correct?
My answer is close (0.12) to the book's answer (0.1). Need to know what the time value should be and why.

Thanks

[Guest]

This question is from Cambridge Checkpoints. I believe that I've already found the answer, using two methods, but do no know which method is technically correct. One or more of the methods could potentially be incorrect by the way. Say something if you think so.

Question: A pole-vaulter of mass 84kg has a 'run-up' speed of 9.1ms/s. What height could his center of mass clear if he clears the bar at a speed of 1.5 ms/s? Neglect the effects of air resistance.

Method #1 : m=84, u=9.1, v=1.5, a=-10, x=?
1.5²=9.1²+(2*-10*x)
1.5²=9.1²-20*x
20x=80.56
x=4.03m

Method #2: Recall that Kinetic energy at the bottom shall be equal to the gravitational potential energy at the top.
84*10*h=3478.02
h=3478.02/840
h=4.14m

What method is correct and why?

[SocialRhubarb]

I can't tell if you've factored it into your working, but for method #2, you don't just have gravitational potential energy at the top, you have some amount of kinetic energy as well.

Both methods should work, although I am somewhat dubious that a run up speed of 9.1 m/s is able to translate directly into some vertical velocity with no loss of speed. I don't particularly like this question.

[Guest]

I can't tell if you've factored it into your working, but for method #2, you don't just have gravitational potential energy at the top, you have some amount of kinetic energy as well.

Both methods should work, although I am somewhat dubious that a run up speed of 9.1 m/s is able to translate directly into some vertical velocity with no loss of speed. I don't particularly like this question.

What you're referring to in your first statement is The total mechanical energy. How does that relate to this question though?

There is a loss of speed. Hence, why 9.1m/s becomes 1.5ms/s at the top, as he clears the bar.
The question itself is simple enough, but what I need to know is when to apply the equations for constant motion if not equations for Energy, in similar scenarios.

[RKTR]

hihi your mistake.. at the top there is still speed but your method 2 assume the speed at the top is 0 ..

so should be mgh=0.5 x 84 x (9.1^2 -1.5^2) 
                     84 x 10 x h =0.5 x 84 x (9.1^2 -1.5^2)
you should also get 4.03m

[SocialRhubarb]

Okay, at the bottom, he has a set amount of energy in the form of kinetic energy. As he jumps some of that kinetic energy is converted into gravitational potential, however, not all of it is converted, as he still retains some of his kinetic energy at the top. The difference in the kinetic energy between when the vaulter is at the bottom and when he is at the top is equivalent to the amount of gravitational potential energy he gains.

.       We can cancel all the 'm's.





Also, the pole vaulter's velocity changes direction, but we assume the magnitude of his velocity is the same. While I might be able to run at 25 km/h, I don't think that while running at 25 km/h, I would be able to convert all of that speed into a vertical velocity of 25 km/h. That was what I was questioning.

[lzxnl]

A few problems that come to mind here.

Firstly, where does the horizontal momentum go? Very unrealistic.

Secondly, an analysis using conservation of energy is incomplete the way you guys have done it. What you've done is work out the height at which the person will have a speed of 1.5 ms^-2 WITH NO ROTATIONAL MOTION.
Now...I don't ever recall seeing a pole vault event where someone clears the bar without rotating their body.

[Guest]

hihi your mistake.. at the top there is still speed but your method 2 assume the speed at the top is 0 ..

so should be mgh=0.5 x 84 x (9.1^2 -1.5^2) 
                     84 x 10 x h =0.5 x 84 x (9.1^2 -1.5^2)
you should also get 4.03m

Ahh right. Makes much more sense. However, I don't get what you've done with your negative values. Isn't acceleration due to gravity -10, since he is jumping upwards and opposing the acceleration of gravity? In order to find the velocity change at the top isn't it 1.5²-9.1²?

Therefore: mgh=0.5 x 84 x (1.5² -9.1²) 
                     84 x -10 x h =0.5 x 84 x (1.5² -9.1²)
Equals 4.03m, but now is the working out how it should be (with appropriate negatives and all)?

EDIT: Some of the negatives weren't equal.