A 5 kg object is raised through a height of 4 m. The Work done by the force of gravity acting on the object is (take g = 10 m/s 2):
–200 J
Calculating Work Done by the Force of Gravity
The question asks us to determine the work done by the force of gravity when a 5 kg object is raised through a height of 4 m. Work done by a force is defined as the product of the force, the displacement, and the cosine of the angle between the force and the displacement.
Understanding Work Done
Work ($W$) is a scalar quantity representing the energy transferred by a force acting over a distance. The formula for work done by a constant force is:
$$W = F \cdot d \cdot \cos(\theta)$$
Where:
- $F$ is the magnitude of the force.
- $d$ is the magnitude of the displacement.
- $\theta$ is the angle between the force vector and the displacement vector.
If the force and displacement are in the same direction, $\theta = 0^\circ$ and $\cos(0^\circ) = 1$, so $W = Fd$. If they are in opposite directions, $\theta = 180^\circ$ and $\cos(180^\circ) = -1$, so $W = -Fd$. If they are perpendicular, $\theta = 90^\circ$ and $\cos(90^\circ) = 0$, so $W = 0$.
Analyzing the Forces and Displacement
In this scenario:
- The object has a mass ($m$) of 5 kg.
- The object is raised through a height (displacement, $d$) of 4 m upwards.
- We need to calculate the work done by the force of gravity ($F_g$).
- The acceleration due to gravity ($g$) is given as 10 m/s$^2$.
The force of gravity acts downwards, towards the center of the Earth. The displacement is upwards, as the object is being raised.
Step-by-Step Calculation of Work Done by Gravity
Step 1: Calculate the magnitude of the force of gravity ($F_g$).
The force of gravity is given by $F_g = m \cdot g$.
$$F_g = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N}$$
Step 2: Determine the displacement ($d$).
The object is raised through a height of 4 m, so the magnitude of the displacement is $d = 4 \, \text{m}$. The direction of displacement is upwards.
Step 3: Determine the angle ($\theta$) between the force of gravity and the displacement.
The force of gravity acts downwards, and the displacement is upwards. These two directions are opposite to each other.
Therefore, the angle between the force of gravity and the displacement is $\theta = 180^\circ$.
Step 4: Calculate the work done by the force of gravity ($W$).
Using the formula $W = F_g \cdot d \cdot \cos(\theta)$:
$$W = 50 \, \text{N} \times 4 \, \text{m} \times \cos(180^\circ)$$
Since $\cos(180^\circ) = -1$:
$$W = 50 \, \text{N} \times 4 \, \text{m} \times (-1)$$
$$W = -200 \, \text{J}$$
Understanding the Negative Work
The work done by the force of gravity is negative (-200 J). This is because the force of gravity acts in the opposite direction to the displacement. When a force opposes the motion or displacement of an object, the work done by that force is negative. In this case, gravity is "trying" to pull the object down while it is being moved up.
Summary of Calculation
| Quantity | Symbol | Value |
|---|---|---|
| Mass | $m$ | 5 kg |
| Acceleration due to gravity | $g$ | 10 m/s$^2$ |
| Force of gravity | $F_g$ | 50 N (downwards) |
| Displacement | $d$ | 4 m (upwards) |
| Angle between $F_g$ and $d$ | $\theta$ | $180^\circ$ |
| Work done by gravity | $W$ | $F_g \cdot d \cdot \cos(\theta) = -200$ J |
Thus, the work done by the force of gravity acting on the object is -200 J.
Revision Table: Key Concepts for Work Done by Gravity
| Concept | Description |
|---|---|
| Work Done | Energy transferred by a force acting over a distance: $W = Fd\cos\theta$. |
| Force of Gravity | $F_g = mg$, acts downwards. |
| Displacement | Change in position of the object. |
| Angle ($\theta$) | Angle between the force vector and the displacement vector. |
| Positive Work | Force and displacement are in the same general direction ($0^\circ \le \theta < 90^\circ$). |
| Negative Work | Force and displacement are in opposite general directions ($90^\circ < \theta \le 180^\circ$). |
| Zero Work | Force is perpendicular to displacement ($\theta = 90^\circ$). |
Additional Information: Potential Energy and Work
When an object is raised against gravity, the work done by the external lifting force is positive and increases the object's gravitational potential energy. The work done by gravity during this upward movement is negative, and it is equal in magnitude but opposite in sign to the change in gravitational potential energy.
The change in gravitational potential energy ($\Delta PE$) when lifting an object of mass $m$ through a height $\Delta h$ is given by:
$$\Delta PE = mg\Delta h$$
In this case, the change in potential energy is $5 \, \text{kg} \times 10 \, \text{m/s}^2 \times 4 \, \text{m} = 200 \, \text{J}$. The work done by gravity is $-\Delta PE$, which is $-200 \, \text{J}$. This confirms our calculation.
Understanding work done by conservative forces like gravity is crucial in physics, particularly in the study of energy conservation.
Similar Questions
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Fill in the blank with the most appropriate option.
The Universal Constant of Gravitation is ________.
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B. Nm2/kg2 is the SI unit of G.
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Important Questions from Gravity
Which of the following law states that, "The force between two objects is directly proportional to the product of their masses?"
Which of the following statements about the movement of planets is true?
A. A planet's orbit is elliptical with the Sun at one of two focal points.
B. The orbit of a planet is circular with the sun in the center.
C. The orbit of a planet is elliptical with another planet in one of the two center-points.
D. The orbit of a planet is circular with another planet in the center.
If the mass of a person is 60 kg on the surface of earth then the same person’s mass on the surface of the moon will be:
The centripetal force required to keep the moon in its orbit is provided by which force?
How is the acceleration due to gravity denoted?
