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Physics 19 Online
OpenStudy (anonymous):

Gravel is dropped on a conveyor belt at the rate of 0.5kg/s. The extra force in Newton required to keep the belt moving at 2m/s, is

OpenStudy (anonymous):

Is it 2/0.5?

OpenStudy (anonymous):

I am not sure.

OpenStudy (anonymous):

are you asking 2 or 0.5?

OpenStudy (anonymous):

Is it 2 metre/second na?

OpenStudy (anonymous):

It's one of the option

OpenStudy (anonymous):

which one? please do say the options.

OpenStudy (anonymous):

(a) 1N (b) 2N (c) 4N (d) 0.5 N

OpenStudy (anonymous):

I think 2 * 0.5 = 1.

OpenStudy (anonymous):

Can you explain it?

OpenStudy (anonymous):

What I think is the conveyor belt is initially at rest. Gravel mass is 0.5 kg. Now given velocity is 2 m / s. So just multiplied using the force formula.

OpenStudy (anonymous):

It is, but you need a reason behind it :D

OpenStudy (anonymous):

I request others to check and criticize me if I am wrong.

OpenStudy (anonymous):

force=momentum*time so momentum is given for unit time (1 second) determine it

OpenStudy (anonymous):

So even my answer is correct I believe.

OpenStudy (anonymous):

the force is nevertheless 0.5N no matter how u speed up

OpenStudy (anonymous):

Momentum is mass*velocity

OpenStudy (anonymous):

But How?

OpenStudy (anonymous):

Abdul, What is the correct option given in the key?

OpenStudy (anonymous):

Key was not given

OpenStudy (anonymous):

Oooooooooooooooo

OpenStudy (anonymous):

sry i dint read the question properly so yes first find momentum and then force

OpenStudy (anonymous):

This was given in FIITJEE

OpenStudy (anonymous):

I could know that. U can even find such questions in BMA workbook.

OpenStudy (anonymous):

In 8th?

OpenStudy (anonymous):

Yeah. Its actually 9th portion. Still u can try.

OpenStudy (anonymous):

Write the net force as: \[\ F_{net}= F - F_{thrust} \] it is -F_thurst because the mass in increasing The net force is zero because the system is not increasing or decreasing its velocity, that is the velocity is constant. Now, \[\ F_{thrust}=v\frac{dm}{dt}\] I think it is fairly easy to solve it now :D.

OpenStudy (anonymous):

Yeah. so my answer is correct.

OpenStudy (anonymous):

What is dm/dt?

OpenStudy (anonymous):

change in mass per unit time or how much mass is added per second.

OpenStudy (anonymous):

Even I will know the use of "d" in formulae in 11th grade. Can anyone explain it in detail?

OpenStudy (anonymous):

d is a symbol for a differential. It represents an infitesimal change of some quantity (you will get to that in calculus). It's like \[\ \Delta \] but infinitely small.

OpenStudy (anonymous):

So, its a very tough concept. What about Calculus? I just know 2 words - Integration and Differentiation.

OpenStudy (anonymous):

Yes that's what you do in Calculus :D. If you are impatient to learn about it I would recommend "Calculus" by Gilbert Strang Link: http://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf

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