The power required to lift a mass of 8⋅0 kg up a vertical distance of 4 m in 2 s is (taking acceleration due to gravity as 10 m/s2):
160 W
Calculating Power Required to Lift a Mass Against Gravity
This problem asks us to determine the power needed to lift a specific mass a certain vertical distance within a given time. Power is defined as the rate at which work is done. To solve this, we first need to calculate the work done against gravity, and then divide it by the time taken.
Understanding Work Done and Power
- Work Done: When we lift an object vertically against gravity, we are doing work against the gravitational force. The work done is calculated as the force applied multiplied by the distance moved in the direction of the force. In this case, the force required to lift the object at a constant speed is equal to its weight.
- Power: Power is how quickly work is done. If a certain amount of work is done in a shorter time, the power required is greater. The formula for power is Work done divided by Time taken.
Given Information
Let's list the values provided in the question:
- Mass of the object (\(m\)) = 8.0 kg
- Vertical distance lifted (\(h\)) = 4 m
- Time taken (\(t\)) = 2 s
- Acceleration due to gravity (\(g\)) = 10 m/s²
Step-by-Step Calculation of Power
First, we calculate the force required to lift the mass. This force must overcome the gravitational force acting on the mass, which is its weight.
The weight of the mass is given by:
Force (\(F\)) = Mass (\(m\)) × Acceleration due to gravity (\(g\))
\(\qquad F = m \times g\)
Substitute the given values:
\(\qquad F = 8.0 \, \text{kg} \times 10 \, \text{m/s}^2\)
\(\qquad F = 80 \, \text{N}\)
Next, we calculate the work done in lifting the mass through the vertical distance.
Work Done (\(W\)) = Force (\(F\)) × Vertical distance (\(h\))
\(\qquad W = F \times h\)
Substitute the calculated force and the given distance:
\(\qquad W = 80 \, \text{N} \times 4 \, \text{m}\)
\(\qquad W = 320 \, \text{J}\)
Finally, we calculate the power required using the formula:
Power (\(P\)) = Work Done (\(W\)) ÷ Time taken (\(t\))
\(\qquad P = \frac{W}{t}\)
Substitute the calculated work done and the given time:
\(\qquad P = \frac{320 \, \text{J}}{2 \, \text{s}}\)
\(\qquad P = 160 \, \text{W}\)
Thus, the power required to lift the mass is 160 Watts.
Summary of Calculation
Here is a quick recap of the steps:
- Calculate the force (weight) needed to lift the mass: \(F = m \times g\).
- Calculate the work done against gravity: \(W = F \times h\).
- Calculate the power required: \(P = W / t\).
Applying the values:
- \(F = 8 \, \text{kg} \times 10 \, \text{m/s}^2 = 80 \, \text{N}\)
- \(W = 80 \, \text{N} \times 4 \, \text{m} = 320 \, \text{J}\)
- \(P = 320 \, \text{J} / 2 \, \text{s} = 160 \, \text{W}\)
The calculated power is 160 W, which matches one of the given options.
Revision Table: Power Calculation
| Concept | Formula | Calculation | Result |
|---|---|---|---|
| Force (Weight) | \(F = m \times g\) | \(8.0 \, \text{kg} \times 10 \, \text{m/s}^2\) | \(80 \, \text{N}\) |
| Work Done | \(W = F \times h\) | \(80 \, \text{N} \times 4 \, \text{m}\) | \(320 \, \text{J}\) |
| Power | \(P = W / t\) | \(320 \, \text{J} / 2 \, \text{s}\) | \(160 \, \text{W}\) |
Additional Information: Power, Work, and Energy
Let's explore some related concepts to deepen our understanding of power, work, and energy in physics.
- Work-Energy Theorem: This theorem states that the net work done on an object is equal to the change in its kinetic energy. In the case of lifting at a constant speed, the change in kinetic energy is zero, meaning the work done by the lifting force is equal and opposite to the work done by gravity.
- Potential Energy: Lifting an object against gravity increases its gravitational potential energy. The increase in potential energy is equal to the work done against gravity (\(mgh\)). The power calculated here is essentially the rate at which potential energy is being increased.
- Units: The standard unit for work is the Joule (J), and the standard unit for power is the Watt (W). 1 Watt is equal to 1 Joule per second (1 W = 1 J/s). The unit of force is Newton (N), and distance is meter (m). \(1 \, \text{J} = 1 \, \text{N} \cdot \text{m}\).
- Power in other contexts: While this problem deals with mechanical power (lifting against gravity), power is a fundamental concept in many areas of physics, such as electrical power (rate of energy transfer in an electric circuit) or thermal power (rate of heat transfer).
Understanding the relationship between force, work, energy, and power is crucial for solving problems involving mechanics and energy transfer.
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