Work question without distance, work and velocity

[MsHomework]

Homework Statement

How much work is done on an 8.0kg wagon rolling along a flat sidewalk, if applied force of 60N opposite to its direction of motion brings it to a rest in 2.0s?

Homework Equations

W = F · d

The Attempt at a Solution

I don't know how to start!

 

[MrB8rPhysics]

Force is the rate of change of momentum... F = (mv-mu)/t

The mass is constant, and the wagon comes to rest.

This should allow you to find the initial velocity of the wagon.

From there either use an equation of motion to find how far it goes, then apply W = f.d, or calculate the kinetic energy of the wagon at its initial velocity - the work done to bring it to rest is equal to that initial KE.

 

[MsHomework]

What does F= (mv-mu)/t stand for? Also the equation for motions would be the five main equations, right?

 

[MrB8rPhysics]

What does F= (mv-mu)/t stand for? Also the equation for motions would be the five main equations, right?

F=(mv-mu)/t is a form of Newtons second law. m for mass, t for time, F for force, v for final velocity, u for initial velocity. Your F here is a force opposing motion so can be used in this formula as -60 N.

Five main equations? I teach them as the 4 SUVAT equations, if that rings a bell...

 

[e.bar.goum]

What does F= (mv-mu)/t stand for? Also the equation for motions would be the five main equations, right?

F = force
m = mass
v = initial velocity
u = final velocity
t = time

It's just the change in momentum caused by stopping the cart.

I'm not sure what you mean by "five main equations", but an equation of motion is simply an equation that relates velocity and/or position and/or acceleration and time. Say, s = ut + 1/2 at^2

EDIT - Snap! Sorry.

 

[MsHomework]

d=(u+v/2)t
u=v+at
d= vt+1/2at^2
u^2=v^2+2ad
d=ut-1/2at^2

 

[netgypsy]

If you make a list of your given variables it makes it a lot easier to see which equation is the simplest to use.

You have force, mass, initial velocity and time. You can go with change in kinetic energy equations or find your distance and use your work equation. Several ways to work it.

 

[MsHomework]

I get a negative intial velocity, is that normal?

 

[e.bar.goum]

Depends on how you set up your problem. Negative values indicate the object is moving in the opposite direction to the direction you set positive to be. (or you made a mistake :D )

 

[MsHomework]

Like I did F= (mv-mu)/t and my initial velocity was -15 m/s

 

[netgypsy]

If you drew your force and velocity vectors what do you notice about their directions? Whether your initial velocity is negative depends totally on which direction you chose as positive, You could be correct or incorrect.

 

[MsHomework]

so the intial velocity is in the opposite direction it is heading?

 

[MsHomework]

Can someone just show me this step by step...because I am lost! This way I could try a similar question to practice this method!

 

[I like Serena]

Hi MsHomework!

Let's take a new approach from scratch.

Suppose the wagon was initially at rest and is accelerated by a force of 60 N for 2 seconds.
First question: what would the acceleration be?

 

[MsHomework]

a= -7.5 m/s^2

 

[MsHomework]

Hi MsHomework!

Let's take a new approach from scratch.

Suppose the wagon was initially at rest and is accelerated by a force of 60 N for 2 seconds.
First question: what would the acceleration be?

But isn' the velocity final at rest since it comes to a rest...?

 

[I like Serena]

a= -7.5 m/s^2

Yes.
I assume you used the formula F=ma to find this?

However, since I was talking about accelerating from rest, let's take the acceleration to be positive shall we?

Next question: what would the speed be after 2 seconds?

 

[I like Serena]

But isn' the velocity final at rest since it comes to a rest...?

Well, I've changed the setting of the problem.
In your problem statement, the wagon is moving along at an unknown speed and is decelerated to a halt.

But, if you'll bear with me, I first want to do a simpler problem, which is the reverse.
That is, we start from rest and then accelerate to the unknown speed.

 

[MsHomework]

I don't have distance!

 

[MsHomework]

alright I will follow your way...so What do i do for distance?

 

[I like Serena]

We'll get to distance in a bit.

Let's first do the speed.
Can you calculate the speed after 2 seconds with an acceleration of 7.5 m/s2?

 

[MsHomework]

15m/s

 

[I like Serena]

Good!
Which formula did you use?

Next question: which distance will the wagon have traveled in these 2 seconds?

 

[MsHomework]

v= a/t... I don't think there is a formula like that...but that is what I used... so ya

 

[MsHomework]

I just divided the acceleration by time to get the speed since the real speed equation requires me to have distance!

 

[I like Serena]

Actually you used: u=v+at
But your formulas are a bit confusing I'm afraid, since the symbols u and v are used inconsistently in your 5 formulas.

Let's hang on to v=at.
Note that there is no division, but multiplication.

So for distance?

 

[I like Serena]

I just divided the acceleration by time to get the speed since the real speed equation requires me to have distance!

Which real speed equation is that?

 

[MsHomework]

v= d/t is the real speed equation .

 

[I like Serena]

v= d/t is the real speed equation .

Not quite, I'm afraid.
This equation is only true if the speed is constant.
In your problem the speed is not constant, so you can't use this equation.Do you have another formula relating distance to acceleration and time?

 

[MsHomework]

velocity final^2= velocity initial^2 + 2(acceleration)(distance)

 

[I like Serena]

velocity final^2= velocity initial^2 + 2(acceleration)(distance)

This one would work, but there is another one with which you can calculate the distance directly.
Can you find it?

 

[MsHomework]

d= 1/2 (velocity initial + velocity final)t

 

[I like Serena]

d= 1/2 (velocity initial + velocity final)t

This is the one formula I can't make sense of.
I can't think of any context where you would want to use it.

Let's not use it.

Do you have yet another one?

 

[MsHomework]

Ok well I decide to just do v=at. then I did velocity final= velocity initial + (acceleration)(distance). Which gave me the initial velocity of 15m/s. Lastly I did the d= 1/2 (velocity initial + velocity final)t equation which gave me a distance of 15 m. is this right?

 

[MsHomework]

Finally after that when I replaced the distance i got into the work equation, I got 900J. Is this right