How to Calculate the Mass of the Left Block in a Pulley System?

[oneamp]

Homework Statement

The figure shows a 100 kg block being released from rest from a height of 1.0 m. It then takes 0.90 s to reach the floor. What is the mass of the block on the left?

Illustration shows a pulley connected to the ceiling, with a block of 100 kg on the right, and a block of unknown mass on the left.

Homework Equations

The Attempt at a Solution

I used this to find acceleration of the system:
0 = 1 + (0.5)(a) [(0.90)^2]
a = (-2.5) m/s^2

I drew a diagram. F_G on the bottom, 100 * 9.8 = 980N down.

F = ma so (100)(2.5) = 250, and that must mean that the vector pointing up in my diagram will be 730N (980 - 250 = 750).

So, the 750 comes from the other block. F = ma again, solve for m I get 292 kg, which is not the right answer.

Where did I go wrong?

Thank you

 

[nasu]

Homework Statement

The figure shows a 100 kg block being released from rest from a height of 1.0 m. It then takes 0.90 s to reach the floor. What is the mass of the block on the left?

Illustration shows a pulley connected to the ceiling, with a block of 100 kg on the right, and a block of unknown mass on the left.

Homework Equations

The Attempt at a Solution

I used this to find acceleration of the system:
0 = 1 + (0.5)(a) [(0.90)^2]
a = (-2.5) m/s^2

I drew a diagram. F_G on the bottom, 100 * 9.8 = 980N down.

F = ma so (100)(2.5) = 250, and that must mean that the vector pointing up in my diagram will be 730N (980 - 250 = 750).

Up to here is (almost) OK.
You found the tension in the rope. It is not clear if it's 730N or 750N (it depends on taking g=9.8 or 10 m/s^2).

So, the 750 comes from the other block. F = ma again, solve for m I get 292 kg, which is not the right answer.

Where did I go wrong?

Thank you

Now write Newton's second law for the other block. You know the tension in the rope and the acceleration (same as for the first block).

 

[Legaldose]

The way I deal with Atwood problems is to first translate the vertical forces to horizontal ones, it makes it easier for me to deal with the forces involved, and gets rid of the confusion of having the pulley. Here's a picture of what I mean. Try applying Newton's Laws to the system on the right.

 

[oneamp]

Thank you :)

 

[Nickie]

for your response. I would approach this problem by first defining the variables and equations that are relevant to the situation. In this case, we have a block of mass 100 kg being released from rest, a pulley, and another block of unknown mass on the left side. We can use the equation F=ma to determine the acceleration of the system, which we know is -2.5 m/s^2 based on the given information.

Next, we can use the equation v=at to find the final velocity of the 100 kg block after it falls 1.0 m. Plugging in the values, we get v=at= (-2.5 m/s^2)(0.90 s) = -2.25 m/s. We can then use this velocity to find the tension in the rope using the equation F=ma, where F is the tension and m is the mass of the block. We know that the tension in the rope is equal to the weight of the block on the left side, so we can set up the equation 980N = 100 kg * (-2.25 m/s^2) and solve for the mass of the block on the left, which comes out to be 43.56 kg.

It seems like you may have made a mistake in your calculation for the tension in the rope, which led to an incorrect mass for the block on the left. Remember that tension is equal to the weight of the block on the left, not the difference between the weight of the block on the right and the force of gravity acting on the block on the right. I hope this helps clarify the problem and leads you to the correct solution.