Calculating Velocity of m1 After Falling a Distance of d Meters over a Pulley

[postfan]

Homework Statement

Two masses, m1 and m2, are hung over a pulley as shown. Assume that m1 is heavier, that the pulley is massless and frictionless, and that the rope does not slip. The blocks are held motionless and then released. Determine the magnitude of the velocity of m1 after it has fallen a distance of d meters. You may enter "m_1" for m1 and "m_2" for m2.

Homework Equations

The Attempt at a Solution

Found the acceleration of the 2 objects, (m_1+m_2)*g, but got stuck after that.

 

[gneill]

You'll have to show your work. Your acceleration doesn't look right.

 

[postfan]

Well the total mass of the 2 objects is (m_1+m_2) and then you multiply it by g.

 

[gneill]

Well the total mass of the 2 objects is (m_1+m_2) and then you multiply it by g.

No, that's not correct (even the units do not yield acceleration... g is an acceleration, and multiplying it by mass gives you force).

Start by drawing the Free Body Diagrams (FBDs) for each mass. There's a force in common provided via the rope.

 

[postfan]

Is the common force tension?

 

[Doc Al]

Is the common force tension?

That's correct.

 

[postfan]

So now what do I do?

 

[Doc Al]

So now what do I do?

You follow gneill's advice. Draw Free Body Diagrams for each mass. What forces act on each?

 

[postfan]

On the block with mass m1 there is a gravitation force of m1*g and a tension force of m2*g, on the block with mass m2 there is a gravitation force of m2*g and there is a tension force of m1*g. Is that right?

 

[Doc Al]

On the block with mass m1 there is a gravitation force of m1*g and a tension force of m2*g, on the block with mass m2 there is a gravitation force of m2*g and there is a tension force of m1*g. Is that right?

No. You have the gravitational forces correct, but the tension is something you'll have to solve for. It's an unknown, so label it "T". (And since the tension is the same throughout the rope, you cannot have it exerting different forces on each mass.)

 

[postfan]

How do you solve for tension?

 

[Doc Al]

How do you solve for tension?

Once you have your FBDs, you'll apply Newton's 2nd law to each mass. That will give you two equations. You'll solve those equations to find the tension, which will be one of the two unknowns.

That's if you want to know the tension. Since the question doesn't ask for it, you can just solve for the acceleration, which is the second unknown.

 

[postfan]

I got g for the acceleration. Assuming it's right, what do I do next?

 

[Doc Al]

I got g for the acceleration. Assuming it's right, what do I do next?

How did you get g for the acceleration? It's not correct. g would be the acceleration of something in free fall, not something attached to ropes and pulleys.

You haven't completed step 1: Draw FBDs for each mass.

 

[postfan]

Yes I did. There are 2 forces acting on each object, gravity and tension.

Using Newtons 2nd I got 2 equations : m1*g-T=m1*a and m2*g-T=m2*a, solving for a gives a=g.

If I did something wrong please tell me what.

 

[Doc Al]

Yes I did. There are 2 forces acting on each object, gravity and tension.

Using Newtons 2nd I got 2 equations : m1*g-T=m1*a and m2*g-T=m2*a, solving for a gives a=g.

If I did something wrong please tell me what.

Ah, you made a sign error in your equation for m2. If m1 accelerates down, then m2 must accelerate up.

(The only way that a = g would be if T = 0. In other words, if the rope were cut and the masses were in free fall.)

 

[postfan]

Ok ,fixing the sign I got (m1-m2)*g/(m1+m2). Is that right?

 

[Doc Al]

Ok ,fixing the sign I got (m1-m2)*g/(m1+m2). Is that right?

Good!

 

[postfan]

Ok , for the next step I used the formula v^2=2*a*x and got v=sqrt(2*g*(m_1-m_2)/(m_1+m_2)*d). Is that right?

 

[Doc Al]

Ok , for the next step I used the formula v^2=2*a*x and got v=sqrt(2*g*(m_1-m_2)/(m_1+m_2)*d). Is that right?

Looks good.

Just for fun, try solving it again using energy methods.