How much work is done when a hoist lifts a 
290-kg
 rock to a height of 
5 m?
 (Use 9.8 m/s2 for the acceleration due to gravity.)

To calculate the work done when lifting a rock, we can use the formula:

\[
\text{Work} = \text{Force} \times \text{Distance}
\]

In this case:

1. **Force** is equal to the weight of the rock, which is given by the product of the mass of the rock and the acceleration due to gravity:

\[
\text{Force} = m \times g
\]

2. **Distance** is the height to which the rock is lifted.

Given:
- Mass (\( m \)) = 290 kg
- Acceleration due to gravity (\( g \)) = 9.8 m/s²
- Height (\( h \)) = 5 m

Now, let's calculate the force:

\[
\text{Force} = 290 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 2842 \, \text{N}
\]

Finally, the work done:

\[
\text{Work} = 2842 \, \text{N} \times 5 \, \text{m} = 14210 \, \text{J}
\]

So, the work done is **14,210 joules (J)**.

Would you like more details or have any questions? Here are five related questions you might find interesting:

1. What is the work done if the mass of the rock was doubled?
2. How does the work change if the height was reduced to 2.5 meters?
3. What if the hoist lifts the rock at a constant speed? Does the work done change?
4. How would the work done be affected if the rock was lifted on the Moon, where gravity is weaker?
5. Can work be done if the rock is lifted vertically but then returns to its original position?

**Tip:** Remember, work is a scalar quantity; it has magnitude but no direction!

How much work is done when a hoist lifts a 290-kg rock to a height of 5 m? (Use 9.8 m/s² for the acceleration due to gravity.)

Learn how to calculate the work done when lifting a 290-kg rock to a height of 5 meters using the work formula: Work = Force × Distance. This problem explains key physics concepts like force, mass, and the work-energy theorem, making it ideal for students in Grades 9-12.

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