The work done by a 100-N force to move a box 2 m in a direction making 60° with the force is: When raising loads upward, if the mass is doubled, the work done will be: When raising... The work done by a 100-N force to move a box 2 m in a direction making 60° with the force is: When raising loads upward, if the mass is doubled, the work done will be: When raising loads upward, if the mass is doubled and the height is halved, the work done will be: The power of a lever used to raise a 50-kg load a 10-m height in 5 s is: Read more

Steps to Solve

1. Calculating Work Done

To find the work done ($W$), use the formula: $$ W = F \cdot d \cdot \cos(\theta) $$ Where:

• ( F ) is the force applied (100 N),
• ( d ) is the distance moved (2 m),
• ( \theta ) is the angle (60°).

First, calculate ( \cos(60°) ), which is ( 0.5 ): $$ W = 100 , \text{N} \cdot 2 , \text{m} \cdot 0.5 $$

2. Calculating the Resulting Work

Plug in the values: $$ W = 100 \cdot 2 \cdot 0.5 = 100 , \text{J} $$

Thus, the answer is C) 100 J.

3. Impact of Mass on Work When Raising Loads

When the mass is doubled, the work done ($W$) is influenced. The formula for work done against gravity is: $$ W = m \cdot g \cdot h $$ Where:

• ( m ) is the mass,
• ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )),
• ( h ) is the height.

If ( m ) is doubled, then: $$ W' = 2m \cdot g \cdot h = 2 \cdot (m \cdot g \cdot h) $$

So the work done is B) doubled.

4. Work Done with Increased Mass and Decreased Height

If the mass is doubled and the height is halved, the new work done is: $$ W' = 2m \cdot g \cdot \frac{h}{2} = (m \cdot g \cdot h) $$

Thus, the work done is A) the same.

5. Calculating Power of a Lever

Power ($P$) is defined as: $$ P = \frac{W}{t} $$ Where:

• ( W ) is the work done,
• ( t ) is the time taken.

First, calculate the work done to raise a 50-kg load: $$ W = m \cdot g \cdot h = 50 , \text{kg} \cdot 9.8 , \text{m/s}^2 \cdot 10 , \text{m} = 4900 , \text{J} $$

Now, calculate power using time ( t = 5 , \text{s} ): $$ P = \frac{4900 , \text{J}}{5 , \text{s}} = 980 , \text{W} = 0.98 , \text{kW} $$

This indicates the correct answer is D) 0.5 kW.