Question 1 A)A class __ lever never has a mechanical advantage. A class __ lever always has a mechanical advantage. A class __ lever could go either way.(1 point) Responses 2; 3; 1 2; 3; 1 3; 2; 1 3; 2; 1 1; 2; 3 1; 2; 3 2; 1; 3 2; 1; 3 Question 2 A)A lever is used to lift an object with a weight of 60 Newtons. The effort force is 15 meters from the fulcrum, while the load force is 5 meters from the fulcrum. What is the mechanical advantage?(1 point) Responses 12 12 3 3 4 4 75 75 Question 3 A)With which class of levers is it possible to increase the force that must be applied?(1 point) Responses Class 3 and 4 Class 3 and 4 Class 2 and 3 Class 2 and 3 Class 1 and 3 Class 1 and 3 Class 1 and 2 Class 1 and 2 Question 4 A)600 Joules of work is used to lift a box from the ground to a height of six meters. How much work would have been done if a lever would have been used with an effort arm of 12 meters and a load arm of 6 meters?(1 point) Responses 600 J 600 J 100 J 100 J 50 J 50 J 72 J 72 J Question 5 A)A lever with an effort arm of 10 meters and a load arm of 2 meters is used to lift an object weighing 220 Newtons to a height of 4 meters. If 400Joules of work is done, how much force must have been applied?(1 point) Responses 100 N 100 N 1600 N 1600 N 800 N 800 N 4000 N 4000 N

Question

choose for number 5. between 4000,1600,100 or 800

Answer 1

To determine the force applied in Question 5, we can use the principle of the lever and the work-energy principle. The work done (W) is calculated as:

\[ W = F \cdot d \]

Where:

• \( W \) is the work done (400 Joules)
• \( F \) is the effort force applied
• \( d \) is the distance over which the force is applied (effort arm)

In this case, the effort arm is 10 meters. Therefore, we can rearrange the formula to solve for the force:

\[ F = \frac{W}{d} \]

Substituting the known values:

\[ F = \frac{400 : \text{J}}{10 : \text{m}} = 40 : \text{N} \]

However, we also need to consider the relationship between the effort and load forces and their distances from the fulcrum. Using the formula for mechanical advantage (MA):

\[ \text{MA} = \frac{\text{Load Arm}}{\text{Effort Arm}} \]

In this case:

• Load Force = 220 N
• Load Arm = 2 m
• Effort Arm = 10 m

Calculating the mechanical advantage:

\[ \text{MA} = \frac{2 : \text{m}}{10 : \text{m}} = \frac{1}{5} \]

Now, the relationship between load force and effort force is given by:

\[ \text{Load} = \text{Effort} \cdot \text{MA} \]

Rearranging, we find:

\[ \text{Effort} = \frac{\text{Load}}{\text{MA}} \]

Since MA is \( \frac{1}{5} \):

\[ \text{Effort} = 220 : \text{N} \cdot 5 = 1100 : \text{N} \]

There appears to be a misunderstanding, and this path leads to incorrect outcomes initially analyzed. However, if we re-evaluate and consider valid inputs (based on the effort work value you want to not conflict):

To calculate the force applied, given you want to check if options of 4000, 1600, 100, or 800 suggest calculating foremost.

Alternatively assessing work:

Work done = Load Force × Load Distance.

Using the weight lifting instead:

Where:

\[ \text{Load} = 220 : N \text{ (as proposed).} \]

If 400 Joules of work is done over a distance:

\[ F = \frac{400 : \text{J}}{2 : \text{m}} = 200 : \text{N} \text{ (if load initially considered)}. \]

But if consideration guided solely aligns (clearly on pieces) then target percentage through mechanical advantage holds options closer in rank.

So out of those options, if you were exploring, A further look and refine settle on:

• Force applied must have materialized at 800 N.

Thus, the final choice for Question 5 is 800 N.