MHB Is this correct?How much work is needed to compress a spring and lift a mass?

[ineedhelpnow]

a mass of 20 kg can be stretched 1 m from the equilibrium position. how much work is needed to compress the spring and lift the mass 1m.

what i did:

$W=F*d=m*g*d$
$=(20 kg)(9.8 \frac{m}{s^2})(1m)=196 J$

is this correct?
a spring on a horizontal surface can be stretched 5m from its equilibrium position with a force of 50 N. how much work is done stretching it 3m from its equilibrium position?

what i did:

$W=50*5=250 J$

then i cross multiply
$5m \to 250 J$
$3m \to W_3$

$W_3=150 J$

correct?

 

[topsquark]

a mass of 20 kg can be stretched 1 m from the equilibrium position. how much work is needed to compress the spring and lift the mass 1m.

what i did:

$W=F*d=m*g*d$
$=(20 kg)(9.8 \frac{m}{s^2})(1m)=196 J$

is this correct?

You need to reword this. Is this is vertical spring? How can you "stretch" a mass??

a spring on a horizontal surface can be stretched 5m from its equilibrium position with a force of 50 N. how much work is done stretching it 3m from its equilibrium position?

what i did:

$W=50*5=250 J$

then i cross multiply
$5m \to 250 J$
$3m \to W_3$

$W_3=150 J$

correct?

How about
[math]F = kx[/math]

[math]50 = k \cdot 5[/math]

Find k, then use [math]W = (1/2)kx^2[/math].

"Cross multiply??"

-Dan

 

[ineedhelpnow]

by a mass it means an object with a mass of 20 kg

and for the second one that would mean $W=0.5*10*3^2=45$

 

[ineedhelpnow]

so it can also be done this way?

$W= \int_{0}^{3} \ 10x dx$

 

[topsquark]

by a mass it means an object with a mass of 20 kg

you wrote:
"a mass of 20 kg can be stretched 1 m from the equilibrium position."
Please explain this sentence!

Is there a spring here? Vertical? Is the equilibrium the point where the mass is hanging stationary on the spring? We need details!

and for the second one that would mean $W=0.5*10*3^2=45$

Yes.

-Dan

- - - Updated - - -

so it can also be done this way?

$W= \int_{0}^{3} \ 10x dx$

And also yes. (Nod)

-Dan

 

[ineedhelpnow]

an object with a mass of 20kg can be stretched 1m from its equilibrium position (idk why we need to know that)? how much work does it take to compress the spring and to lift the mass UP 1m.

 

[topsquark]

an object with a mass of 20kg can be stretched 1m from its equilibrium position (idk why we need to know that)? how much work does it take to compress the spring and to lift the mass UP 1m.

Is this the actual problem as it has been given to you? It stinks.

Let me try to re-word this to make some kind of sense. If it's not your given problem, just let us know.

A vertical spring with spring constant k has a 20 kg mass hanging from it such that the spring is stretched 1 m from its equilibrium length when the mass hangs freely. How much work is done to lift the mass 1 m above the equilibrium position?

We need k. We know that the object is hanging in place at the start, so we know that mg = F_s. This will give you k.

Now use [math]W = (1/2)kx^2 - (1/2)kx_0^2[/math] to find the work.

-Dan

 

[ineedhelpnow]

that's how the problem was given. he actually brought the exact same question on the test. turns out the answer was 294 J :/

 

[topsquark]

Hmmm...And that isn't the solution to the problem I wrote for you. I have no idea what the problem should look like.

Kick your instructor in the shins for me.

-Dan

 

[Dethrone]

It's been a long time since I did physics, but I'll give it a try.

So it's a badly written question, but I think I understand what it wants.
As Dan says, work is given by the following:

$$W = \frac{1}{2}kx^2$$, but we need to first determine $$k$$. That is simple, by equating the gravity force with the spring force.

$$(20)(0.8)=k(1)$$, k = 196. Plugging that back into the equation for work, we can see that the work is 98 J.

So the work needed to lift the mass back to its equilibrium position, and then 1 more meter up is:
$$
W = F_s + F_g = 98 + mgd = 98 + 20(9.8)(1) = 294 J$$