Change in velocity due to force acting over given time

[Giu1iano]

What change in velocity would be produced by an unbalanced force of 2.4 × 10^4 N acting for 6 s on a 2.0 × 10^3 kg dragster?

Not sure how to setup this problem Equations i have are as follows:

Fnet = mass × acceleration
Velocity2 = velocity1 + acceleration × time
Distance = velocity1 × time + 1/2acceleration × time ^2

d=((v1 + v2) ÷ 2)t
v2^2=v1^2+2ad

I don't know how to attempt this question and my teachers hasn't answered my email.

Please help

 

[Simon Bridge]

Hint: change in velocity is given by ##\Delta v = v_2-v_1##
Start by listing what you know: i.e. assign the variables in the equations to the corresponding numbers in the problem.

 

[Let'sthink]

Because time 6 s is given only, "Velocity2 = velocity1 + acceleration × time" this formula is enough.
calculate 'a' first; a = [(2.4 × 10^4 )/(2.0 × 10^3)] = 12 m/s^2
required change in velocity = (Velocity2 - velocity1) = 12×6 = 72 m/s

 

[Giu1iano]

Because time 6 s is given only, "Velocity2 = velocity1 + acceleration × time" this formula is enough.
calculate 'a' first; a = [(2.4 × 10^4 )/(2.0 × 10^3)] = 12 m/s^2
required change in velocity = (Velocity2 - velocity1) = 12×6 = 72 m/s

How does 24000N/2000kg = 12 m/s^2?

 

[Simon Bridge]

How does 24000N/2000kg = 12 m/s^2?

... hint: do you have an equation with force and mass in it?

Note: we do not usually do your homework for you.

 

[Giu1iano]

How does 24000N/2000kg = 12 m/s^2?

Never mind.

 

[Simon Bridge]

Do you understand how to do the problem now?

 

[berkeman]

Because time 6 s is given only, "Velocity2 = velocity1 + acceleration × time" this formula is enough.
calculate 'a' first; a = [(2.4 × 10^4 )/(2.0 × 10^3)] = 12 m/s^2
required change in velocity = (Velocity2 - velocity1) = 12×6 = 72 m/s

Just a reminder -- do not do student's work for them. Please just provide hints, ask probing questions, find mistakes, etc. But it is against the PF homework rules to do their work for them. Makes sense?

 

[Kaura]

You could use the equation Impulse = Force x Time to find the change of momentum and subsequently the change in velocity

 

[Giu1iano]

Do you understand how to do the problem now?

I'm so tired I 2000kg as 2kg, and yes understand how to solve the problem.

Thank you!

 

[Simon Bridge]

You could use the equation Impulse = Force x Time to find the change of momentum and subsequently the change in velocity

Nice lateral thinking...
That approach requires the concept of "specific impulse" and 3 additional equations though, vis: ##\Delta p = m\Delta v## and ##I=Ft## and ##I=\Delta p## ... all that simplifies down to ##F=ma## which OP already has. But sure: that would work and showing various approaches can be helpful.

I'm so tired I 2000kg as 2kg, and yes understand how to solve the problem.

... well done then.

Just so others benifit more:
The general approach to a problem like this is to start by drawing a diagram, labelling it, then listing what you know in terms of the symbols you are going to use.
In the above case you know:
F=24000N, m=12000kg, and Δt=6s ... you need Δv=? (check the units are consistent)

Now you want a bunch of equations

...by definition Δv=aΔt (constant acceleration) but that has two unknowns.
You want to find Δv so you need an equation with ##a## in it and F=ma sprang to mind your mind with the others.
Use algebra to make an equation with what you want on the left and everything you know on the right. In general you may need more than two equations.

In this case it is a matter of solving the simultaneous equations:
##F=ma \implies a=F/m## sub into ##\Delta v = a\Delta t## to give ##\Delta v = F\Delta t /m##

The last step is plugging the numbers into the equations: ##\Delta v = 24000\times 6 \div 12000 = 72##m/s

You can also do the number-crunching in two steps, finding the acceleration first.
In general it is best practise to do all the algebra before you crunch numbers.

The toughest part of this approach is finding the equations - the usual mistake is to try to do this by memorizing lots of equations. The best practise is to remember the physics that gives rise to the equations. In this case, you need to understand what "acceleration" means. Then writing down the equation is just a matter of using the maths as a language.

 

[Let'sthink]

I am replying to the specific sub question of the original asker. How (24000 N)/(2000 kg)
I hope you know 24000?2000 = 12 Now only you have to expand N into fundamental units Kg m and s, which comes from the the relation F = ma
[N] = [(kg×m)/s^2] Hence [N]/[kg] = [m/s^2] You need to repeat this procedure in many other problems to understand it also frame questions your self such as why [J] = [Nm] and also = [(kg m^2)/s^2], Which defining relations in Physics give us these equivalent expressions for [J]. You have a feeling that one has to study and then solve problems. I tell you one more thing study and generate problems to answer.