# Calculating Work in a Pulley System

• james brug
In summary, the student attempted to solve for the force on the 20N block, but found that it was more complicated than originally thought. They then found that the work done on the other block would not be the same.
james brug
Work problem--blocks,pulley

## Homework Statement

Two blocks are connected by a very light string passing over a massless and frictionless pulley . The 20.0 N block moves 75.0 cm to the right and the 12.0 N block moves 75.0 cm downward.

Find the total work done on 20.0N block if there is no friction between the table and the 20.0N block.

## Homework Equations

$$w=f\cdot s$$
$$f_k=\mu_k\cdot n$$

## The Attempt at a Solution

w=12N(.75m)=9J--mastering physics says it's wrong.

Ya, your attempt is not correct. What you need to do is find out the force the hanging block exerts on the sliding block. Then you can do W=F*d using this force over a distance of 75cm.

First, are you sure the blocks are in Newtons? Very odd - would expect kg.
Second, I think the whole thing is more complicated because the blocks are accelerating. A force of 12 N down on the 2nd block will be opposed by the Tension in the string, so
sum of forces = ma
mg - T = ma
T = mg-ma and it will not be the full 12N pulling on the other block.
I think the first step will be to calculate the acceleration of the whole system.

Second, I think the whole thing is more complicated because the blocks are accelerating. A force of 12 N down on the 2nd block will be opposed by the Tension in the string, so
sum of forces = ma

That's probably it. There is a problem nearly identical to this in the book, but it says the blocks are moving at a constant speed.

I believe Delphi51 is correct.

T would give me the force on the 20N block and thus the work, yes?
a=12N/total weight?

I think you mean that a=(Force)/(mass) not weight. For the force, use the T from your first statement.

What first statement?

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james brug said:
T would give me the force on the 20N block and thus the work, yes?

This statement. You say that solving for T would give you the force you need in the equation to find work.

What about the work done on the other block? Is it the same?

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The work will not be the same for the other block. In this case the only force acting on it is mg. Therefore, your work will just be mg*.75m and I think it will be negative.

## 1. What are work problem-blocks and pulleys used for?

Work problem-blocks and pulleys are used to make tasks easier by reducing the amount of force needed to lift or move objects. They are commonly used in machines and systems for transportation, construction, and other industries.

## 2. How do work problem-blocks and pulleys work?

Work problem-blocks and pulleys use the principle of mechanical advantage to reduce the amount of force needed to move an object. By distributing the weight of the object over multiple pulleys or changing the direction of the force, less force is required to lift or move the object.

## 3. What are the different types of pulleys used in work problems?

There are three main types of pulleys used in work problems: fixed pulleys, moveable pulleys, and compound pulleys. Fixed pulleys have a stationary axle and are used to change the direction of the force. Moveable pulleys have a moveable axle and are used to reduce the amount of force needed. Compound pulleys combine fixed and moveable pulleys to create an even greater mechanical advantage.

## 4. What are some examples of work problems that use pulleys?

Some examples of work problems that use pulleys include lifting heavy objects, hoisting sails on a boat, and operating cranes at construction sites. Pulleys can also be found in exercise machines, elevators, and even in simple tools like a clothesline.

## 5. How do you calculate the mechanical advantage of a pulley system?

The mechanical advantage of a pulley system is calculated by dividing the load by the effort (MA = Load/Effort). For example, if a pulley system requires a force of 50 pounds to lift a 100-pound object, the mechanical advantage would be 100/50 = 2. This means that the pulley system reduces the amount of force needed by half.

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