1. ## Another physics-like question...

A particle mass 2kg on a smooth level surface and is at rest. A horizontal force of 100 Newtons acts on the particle. No. of seconds it will take for particle to be moved 10 m is?

I've drawn a force diagram. mg = 2g acting down. N acting up. 100N acting sideways. Then what do I do? Thx

2. Originally Posted by classicstrings
A particle mass 2kg on a smooth level surface and is at rest. A horizontal force of 100 Newtons acts on the particle. No. of seconds it will take for particle to be moved 10 m is?

I've drawn a force diagram. mg = 2g acting down. N acting up. 100N acting sideways. Then what do I do? Thx
I'm glad to hear you drew the full Free-Body Diagram. (Few do!) The vertical forces, of course, do nothing here since $\displaystyle a_{vert} = 0$ implies $\displaystyle \sum F_{vert} = ma_{vert} = 0$, but it's a good habit to always quickly check this.

When in doubt use Newton's 2nd. Here it is simple, since there is only one horizontal force. So $\displaystyle \sum F= ma$ implies $\displaystyle a = F/m$ = 100 N/2 kg = 50 m/s^2.

So we know that the particle starts from rest and has an acceleration of 50 m/s^2 and moves for 10 m. (For the record I am setting up a +x direction to be in the direction of the acceleration and an origin to be at the point where the particle was at rest.)

The quickest equation to use would be:

$\displaystyle x = x_0 + v_0t + (1/2)at^2$

$\displaystyle 10 = (1/2)(50)t^2$

$\displaystyle t = \sqrt{\frac{2 \cdot 10}{50}} = 0.63246 \, s$

Or t = 0.63 s to 2 significant figures.

-Dan

3. Through the questions that you have taken time to do for me - F = ma is the key point always like you said. Got to keep that in mind, then use those constant acceleration formulae - s = ut + 0.5at^2 , v = u + at , v^2 = u^2 + 2as

4. Originally Posted by classicstrings
Through the questions that you have taken time to do for me - F = ma is the key point always like you said. Got to keep that in mind, then use those constant acceleration formulae - s = ut + 0.5at^2 , v = u + at , v^2 = u^2 + 2as
There's another that's occasionally useful as well:
$\displaystyle s = \frac{1}{2}(u + v)t$ (in your notation)

Once in a great while you'll get a problem where you don't have the acceleration, though you know it's constant. This equation uses the average velocity to find displacement.

-Dan

5. Ah ok.. so for example this question...

Car travel at 25m/s when brakes applied. Brought to rest with deceleration in 3seconds. How far will it travel after brakes applied?

u = 25 m/s
t = 3
v = 0
s = ?

s = (1/2)(u + v)t
= 0.5 * 75
= 37.5 metres

However, I am wondering how you work this out if you didnt know the formula...

Ah - then you work out the acceleration....

v = u + at
0 = 25 + 3a
a = -25/3 m/s^2

Then plug in v^2 = u^2 + 2as

0 = 625 + 2*(-25/3)(S)
s = 37.5 metres

....

6. Originally Posted by classicstrings
Ah ok.. so for example this question...

Car travel at 25m/s when brakes applied. Brought to rest with deceleration in 3seconds. How far will it travel after brakes applied?

u = 25 m/s
t = 3
v = 0
s = ?

s = (1/2)(u + v)t
= 0.5 * 75
= 37.5 metres

However, I am wondering how you work this out if you didnt know the formula...

Ah - then you work out the acceleration....

v = u + at
0 = 25 + 3a
a = -25/3 m/s^2

Then plug in v^2 = u^2 + 2as

0 = 625 + 2*(-25/3)(S)
s = 37.5 metres

....
Bingo!

The advantage of the set of four equations over the set of three that you listed is that occasionally (as above) you will need to use two equations to get to your answer if you only have the three equations. Even more rarely (though it comes up often in 2 and 3-D motion) is if you have only use the three equations you might have to solve simultaneous equations. (Which, I suppose, is technically happening when you have to solve two equations instead of one.)

-Dan