Question

A 50kg box is lifted 0.5m by a rope. If 300J of work is done, what is the speed of the box at the highest point?

Answer 1

To determine the speed of the box at the highest point, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

Let's break down the steps:

1. **Calculate the gravitational potential energy gained** as the box is lifted:
\[
\text{Potential Energy (PE)} = m \cdot g \cdot h
\]
Where:
- \( m = 50 \, \text{kg} \) (mass of the box)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 0.5 \, \text{m} \) (height lifted)

Plugging in the values:
\[
\text{PE} = 50 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 0.5 \, \text{m} = 245.25 \, \text{J}
\]

2. **Determine the net work done** on the box. The total work done on the box is given as 300J. The net work done will be the work done minus the potential energy gained:
\[
\text{Net Work} = \text{Total Work} - \text{Potential Energy}
\]
\[
\text{Net Work} = 300 \, \text{J} - 245.25 \, \text{J} = 54.75 \, \text{J}
\]

3. **Use the net work to find the change in kinetic energy**:
\[
\text{Net Work} = \Delta KE = \frac{1}{2} m v^2
\]

4. **Set up the equation**:
\[
54.75 \, \text{J} = \frac{1}{2} \cdot 50 \, \text{kg} \cdot v^2
\]

5. **Solve for \( v^2 \)**:
\[
54.75 = 25 v^2 \implies v^2 = \frac{54.75}{25} = 2.19
\]

6. **Take the square root to find \( v \)**:
\[
v = \sqrt{2.19} \approx 1.48 \, \text{m/s}
\]

Thus, the speed of the box at the highest point is approximately \( 1.48 \, \text{m/s} \).