Deceitful Wing's question thread

[Deceitful Wings]

Please help!
My questions could help others!

[Phy124]

Assuming it is shot out horizontally and taking g = 10 m/s²

Question 11:

Vertically;











12.

Horizontally;





edit: added units, can't forget those

[Deceitful Wings]

thanks!

[Deceitful Wings]

How does a pn junction work? The explanation in my textbook sucks!

[Deceitful Wings]

A ball of mass 150 g is launched from a spring when it is released from a compressed state. The ball accelerated from rest. Whilst in contact with the spring, the force on the ball decreases uniformly from 120 N to 0 N in a time of 20ms. The ball rolls up the incline and into the box. You may assume that the collision between the ball and spring is elastic. (See attached image)

a) What is the speed of the ball immediately after it loses contact with the spring?

b)The spring constant is 375 N/m, by how much was the spring initially compressed?

c) If we consider all the surfaces to be frictionless, (floor and inclined ramp), will the ball make it up the ramp? Justify your answer with calculations.

d) What is the minimum constant friction force the ramp must apply to the ball to ensure that the ball does not reach the top?

[DisaFear]

Edited, because DisaFear is dumb and EastsideR is smart

a) We know that Impulse (symbol: J) is equal to Change in Momentum
J = 120N * 0.02s = 2.4 Ns (It is 1.2N)
This is the change in momentum of the ball.
Change in momentum = final momentum - initial momentim (zero in this case)
1.2 = 0.150 * v - 0
v = 8m/s

b) We know from Hooke's Law that restoring force is equal to spring constant times distance of compression/extension, shown as F=-kx
So in this case, F=120N, k=375N/m, compression = ?
120N = 375N/m * x
Compression = 0.32m

c) Used an energy approach for this
Let's look at the total energy in this system. By working out the initial KE, KE=0.5*m*v^2, we find that the kinetic energy comes to 4.8J
The surface is frictionless. Me thinks, this ball will go on forever horizontally, right? So now, to change its height from 0m to 2.7m, how much energy do we need?

U(g) = mgh = 0.150 * 10 * 2.7 = 4.05J

We have more than enough energy to get up the hill. If you wanted to calculate the speed of the ball after it climbs the hill for any reason, it turns out to be 3.2m/s, which seems reasonable.

d) (I am least sure on this question) Work done = Change in Kinetic Energy, according to Work-Kinetic Energy Theorem.
We need to do some work on this system to take energy out of the system, so it does not reach the top. With me?

The total energy in the system is 4.8J
To reach the top, we need 4.05J

Work done = Final KE - Initial KE = 4.05J - 4.8 = 0.75 (We need to do negative work to remove energy from the system)
The friction vector is directly opposing direction of motion
W = F * change in distance = 4.2m (the hypotenuse of the triangle) * F
F = -0.18N (Force is a vector, the negative sign means it is in the direction opposing motion)

^I think I've done something wrong here, with the angles possible. I'll look over it. But it should give you some sort of direction to carry on from. Someone correct me if I'm wrong

^EastsideR says its right, so should be

(Do you have answers to this problem?)

[SenriAkane]

a) We know that Impulse (symbol: J) is equal to Change in Momentum
J = 120N * 0.02s = 2.4 Ns

You missed the statement "The force decreased Uniformly". Draw a F-t graph and change in momentum should be 1.2 Ns

This is the change in momentum of the ball.
Change in momentum = final momentum - initial momentim (zero in this case)
2.4 = 0.150 * v - 0
v = 16m/s

b) We know from Hooke's Law that restoring force is equal to spring constant times distance of compression/extension, shown as F=-kx
So in this case, F=120N, k=375N/m, compression = ?
120N = 375N/m * x
Compression = 0.32m

c) Used an energy approach for this
Let's look at the total energy in this system. By working out the initial KE, KE=0.5*m*v^2, we find that the kinetic energy comes to 19.2J
The surface is frictionless. Me thinks, this ball will go on forever horizontally, right? So now, to change its height from 0m to 2.7m, how much energy do we need?

U(g) = mgh = 0.150 * 10 * 2.7 = 4.05J

We have more than enough energy to get up the hill. If you wanted to calculate the speed of the ball after it climbs the hill for any reason, it turns out to be 14.2m/s, which seems reasonable.

d) (I am least sure on this question) Work done = Change in Kinetic Energy, according to Work-Kinetic Energy Theorem.
We need to do some work on this system to take energy out of the system, so it does not reach the top. With me?

The total energy in the system is 19.2J
To reach the top, we need 4.05J

Work done = Final KE - Initial KE = 4.05J - 19.2J = -15.15 (We need to do negative work to remove energy from the system)
The friction vector is directly opposing direction of motion
W = F * change in distance = 4.2m (the hypotenuse of the triangle) * F
F = -3.6N (Force is a vector, the negative sign means it is in the direction opposing motion)

^I think I've done something wrong here, with the angles possible. I'll look over it. But it should give you some sort of direction to carry on from. Someone correct me if I'm wrong

(Do you have answers to this problem?)

[Ndon95]

Could you please answer this ..
please check attachment for image .
edit:forgot to add whats below..
A cyclist pedals up a 150  slope at a constant speed as shown.  The total mass of rider and bicycle is 100 kg.

Question 4
Which of the speeds (A–D) is the average speed of the rider for the whole journey? You may ignore the time taken to turn around.
A. 5.3 ms-1      B. 5.7 ms-1      C. 6.0 ms-1      D. 6.3 ms-1

[DisaFear]

A cyclist pedals up a 150  slope at a constant speed as shown.  The total mass of rider and bicycle is 100 kg.

Is that a 150m?

[Lasercookie]

Could you please answer this ..
please check attachment for image .
edit:forgot to add whats below..
A cyclist pedals up a 150  slope at a constant speed as shown.  The total mass of rider and bicycle is 100 kg.

Question 4
Which of the speeds (A–D) is the average speed of the rider for the whole journey? You may ignore the time taken to turn around.
A. 5.3 ms-1      B. 5.7 ms-1      C. 6.0 ms-1      D. 6.3 ms-1

Is that a 150m slope?

And what's turning around got to do with it? Was there more information to the question?

[Bhootnike]

Could you please answer this ..
please check attachment for image .
edit:forgot to add whats below..
A cyclist pedals up a 150  slope at a constant speed as shown.  The total mass of rider and bicycle is 100 kg.

Question 4
Which of the speeds (A–D) is the average speed of the rider for the whole journey? You may ignore the time taken to turn around.
A. 5.3 ms-1      B. 5.7 ms-1      C. 6.0 ms-1      D. 6.3 ms-1

Is that a 150m slope?

And what's turning around got to do with it? Was there more information to the question?

looking at the diagram given , maybe he meant 15 degrees, not 150 m - but 150m could work.. although time isnt given :S

Ok. i spent a while doing this. doesnt seem possible with out some more values. haha... :s


but yer, lets see if he meant 15 degrees!:

it seems as if there was a time value given in the question before this one, - which might be useful in the calculation of v.

i think, but im not sure, cause it says constant speed.. , that you do:
find horizontal component of W=mg,i.e 1000cos15 = 972.3699...
Direction of moving force is to the right, so the force to the left must also be 972.369 - since theres constant speed. (irrelevant to answer,just saying..)

~ what i'm not sure about is that if you were to apply f = ma, you'd get an acceleration value of 9.72 which does not seem right since its explicitly stated constant speed.

so f = ma
then a = 9.72..m/s^2

some straight line motion equation, we need to figure out speed.

mm. we're missing a distance, or a time value here.

so maybe if he did mean 150m.
d = 150
u=0
v=?
a=9.72

v= root 2ax
v= root 2*9.62*150
v= ...... no. never. impossible.

[Ndon95]

Sorry about the confusion its 15 degrees stupid word wont copy it as degrees ...
and ahha thats all the values given A cyclist pedals up a 15 degrees  slope at a constant speed as shown.  The total mass of rider and bicycle is 100 kg.

Question 1
What is the magnitude of the net force on the bicycle and rider when travelling up the slope?

Question 2
Calculate the magnitude of the normal reaction force acting on the bicycle when travelling up the slope.

Question 3
Calculate the magnitude of the total frictional forces that are acting on the rider and bicycle while travelling down the slope.

Question 4
Which of the speeds (A–D) is the average speed of the rider for the whole journey? You may ignore the time taken to turn around.
A. 5.3 ms-1      B. 5.7 ms-1      C. 6.0 ms-1      D. 6.3 ms-1

those are all the questions

[Bhootnike]

do you know the answer by any chance? is it c?

[Ndon95]

ye the answer is a 

[Bhootnike]

ye the answer is a 

i dont know man! thats hard!  unless im missing out on something obvious
a time value wouldve been nice. and a distance!

Laser/disa/anyone else, you guys got any idea?

all i got to was that, 1000sin15 = 258.8 N, so using f=ma, a = 2.588.