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Anonymous Fri 16 Nov 18:21:12 2018 No.10145521 View Reply Original Report

Quoted By: >>10145542

Can someone tell me what equations I can use to solve an exercise saying: "A man is pushing a cart of 100kg horizontally with the speed of 1m/s. If he changes the force of pushing by delta F=20N, then what will be the speed 3s after increasing the force?

I dont even know what to begin studying to solve this shit.

Anonymous

Anonymous Fri 16 Nov 2018 18:26:44 No.10145531 Report

Quoted By: >>10145681

$F = ma \qquad v = u + at$
$v = u + \bigg(\frac{F}{m}\bigg)t$
$v = 1 + \bigg(\frac{20}{100}\bigg)3$
$v = 1.6 \quad ms^{-1}$

Anonymous

Anonymous Fri 16 Nov 2018 18:29:58 No.10145537 Report

Quoted By: >>10145681

F = m * a
=> a = F / m = 20N / 100kg = 0.2 m/s^2
v = a * t = 0.2m/s^2 * 3s = 0.6 m / s
v = v1 + v2 = 1.6 m/s

Anonymous

Anonymous Fri 16 Nov 2018 18:31:18 No.10145541 Report

Quoted By: >>10145681

I guess it works somewhat like that:
p_0=mv (=100 kgm/s)
F=dp/dt
applying the force for 3 seconds adds an impuls:
F*dt=dp_1
F*3=p_1
p=p_0+p1=160kgm/s
v=p_1/m=1.6m/s

don't really know though

Anonymous

Anonymous Fri 16 Nov 2018 18:32:22 No.10145542 Report

Quoted By: >>10145681

>>10145521
Acceleration from Newton's seconds law. Final speed from kinematics.

Anonymous

Anonymous Fri 16 Nov 2018 18:49:16 No.10145582 Report

Quoted By: >>10145681

okay so it is going 1 m/s then it speeds up by 20 newtons of force so just check how much 20 newtons of acceleration is on a 100kg object it is

F=m*a,
F/m=a
it's .2ms/s

so every second it increases speed by .2 over three seconds it is .6 add that to 1 second and get 1.6ms

Anonymous

Anonymous Fri 16 Nov 2018 19:35:32 No.10145681 Report

Quoted By: >>10146679

>>10145582
>>10145542
>>10145541
>>10145537
>>10145531
What the fuck. Didn't realize it was this simple. This is sad... thanks

Anonymous

Anonymous Sat 17 Nov 2018 02:14:52 No.10146679 Report

Quoted By:

>>10145681
>What the fuck. Didn't realize it was this simple. This is sad... thanks

Protip: Try to derive the kinematic equations from the definitions of velocity and aceleration on your own. You will never have any problems with exercises of this kind ever again.

https://www.carolina.com/teacher-resources/Interactive/derivation-of-the-kinematics-equation/tr32615.tr

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