What Is the Velocity of Block M After Mass m Falls 2.5 Meters?

[Selerinus]

I've been struggling over this homeword question for hours now, and have made little, or no progress figuring out the exact steps for it.

The system is set up with block M, on a table, attached on the left side of the block is a string, which is holding, across a pulley, little mass m, which dangles over the edge of the table. The exact difference in angles would be 90'.

In the system shown, block M (15 kg), is initially moving to the left, with a speed v(i) = 2.8 m/s. The mass of m, which dangles in the air is 8 kg. There is no mass in the string, and no friction in the pulley. The coefficent of friction betweek M and the surface are kinetic friction = .3, and static friction = .4.

Find the velocity of M when m has fallen 2.5 m

Any help would be appreciated.

 

[OlderDan]

I've been struggling over this homeword question for hours now, and have made little, or no progress figuring out the exact steps for it.

The system is set up with block M, on a table, attached on the left side of the block is a string, which is holding, across a pulley, little mass m, which dangles over the edge of the table. The exact difference in angles would be 90'.

In the system shown, block M (15 kg), is initially moving to the left, with a speed v(i) = 2.8 m/s. The mass of m, which dangles in the air is 8 kg. There is no mass in the string, and no friction in the pulley. The coefficent of friction betweek M and the surface are kinetic friction = .3, and static friction = .4.

Find the velocity of M when m has fallen 2.5 m

Any help would be appreciated.

There are a couple of approaches to this problem. One approach will permit you to find the tension in the string, the other will not. Since you are not asked for the tension, you can consider it to be an "internal force" and treat the two blocks as one object. You can do the problem as if both masses were on the table (if you assume no friction acts on the small mass) and you were pulling the little mass horizontally with a force equal to its weight. The string and pulley only serve to change the directions of motion. Both masses move the same distance with the same speed and the same magnitude of acceleration.

 

[Ja4Coltrane]

there are several ways to solve this problem, but since it gave you distance for mass m, I would attack it with an energy approach.

you want to find the acceleration of the system with no friction. Multiply by mass M to get the force acting on mass M excluding friction (what is this force by the way)? Now it's easy. Work=force *distance and work is the change in kinetic energy, right?

Don't forget about friction and the initial speed!

 

[Selerinus]

So how do the forces of kinetic and static friction factor in. I've already tried solving it, every which way I can. The actual mass of the object seems independent of the system, because it seems to me, that they cancel out.

Solving for them as one mass doesn't seem to work, as I keep getting a velocity way off.

K(i) = K(f) + U(f)

I divided out the masses, and solves for the velocity, but its way too high.

v(i)^2 = v(f)^2 + 2gh

(2.8 m/s)^2 = v(f)^2 + 2 (9.8 m/s^2)(2.5m)
v(f) = about 6 m/s, but the book shows the answer more at about 3.

I assumed that the acceleration of the object would be dependent on gravity, since the dangling mass m, has a mass of 8 kg, and a force of 9.8 m/s^2 pulling down on it
So the force pulling down on it, and subsequently on M, is 78.4N, but I can't figure out where to go from here, except that I have already connected on the concept of the force, and such, not really being dependent on the angle.

 

[OlderDan]

So how do the forces of kinetic and static friction factor in. I've already tried solving it, every which way I can. The actual mass of the object seems independent of the system, because it seems to me, that they cancel out.

Solving for them as one mass doesn't seem to work, as I keep getting a velocity way off.

K(i) = K(f) + U(f)

I divided out the masses, and solves for the velocity, but its way too high.

v(i)^2 = v(f)^2 + 2gh

(2.8 m/s)^2 = v(f)^2 + 2 (9.8 m/s^2)(2.5m)
v(f) = about 6 m/s, but the book shows the answer more at about 3.

My mistake. I lost track of the friction. I will go back and correct my post. You can treat the small mass as I said if you assume it is frictionless.

You have taken a completely different approach to the problem, using energy instead of just looking at forces. This will work if you include the work done by friction in the calculations. Since the mass is moving initially, and continues to move, only kinetic friction is relevant. Can you compute the work done by friction? Can you incorporate that into your equation?

You need to be careful with the masses. Mass is not going to divide out of a correct energy equation.

 

[Selerinus]

Woo, got it. Thanks for the help.