A particle of mass 40 g is thrown vertically upwards with a speed of 10 ms -1 . Find the work done by the force of gravity during the time the particle goes up.
-2.0 J
Calculating Work Done by Gravity on a Vertically Moving Particle
Let's analyze the problem of a particle moving vertically upwards under the influence of gravity and calculate the work done by the gravitational force during its upward journey.
Understanding Work Done by a Force
Work done by a constant force is defined as the product of the force, the displacement, and the cosine of the angle between the force and the displacement vectors. Mathematically, work done (W) is given by:
\(W = F \cdot d \cdot \cos(\theta)\)
Where:
- \(F\) is the magnitude of the force.
- \(d\) is the magnitude of the displacement.
- \(\theta\) is the angle between the force and displacement vectors.
Given Information
We are given the following information about the particle:
- Mass of the particle, \(m = 40 \, \text{g}\).
- Initial velocity, \(u = 10 \, \text{ms}^{-1}\) (upwards).
We need to find the work done by the force of gravity while the particle goes up. The force of gravity acts downwards. The displacement of the particle as it goes up is in the upward direction.
Step-by-Step Calculation of Work Done by Gravity
Step 1: Convert Units
The mass is given in grams. We need to convert it to kilograms (kg) for standard SI units.
\(m = 40 \, \text{g} = 40 \times 10^{-3} \, \text{kg} = 0.040 \, \text{kg}\)
Step 2: Determine the Force of Gravity
The force of gravity acting on the particle is given by \(F_g = m \cdot g\), where \(g\) is the acceleration due to gravity. We usually take \(g \approx 9.8 \, \text{ms}^{-2}\).
\(F_g = 0.040 \, \text{kg} \times 9.8 \, \text{ms}^{-2} = 0.392 \, \text{N}\)
This force acts vertically downwards.
Step 3: Determine the Displacement (Maximum Height Reached)
The particle goes up until its final velocity becomes zero at the maximum height. We can use the kinematic equation:
\(v^2 = u^2 + 2as\)
Where:
- \(v\) is the final velocity (\(0 \, \text{ms}^{-1}\) at max height).
- \(u\) is the initial velocity (\(10 \, \text{ms}^{-1}\)).
- \(a\) is the acceleration. While going up, the acceleration is due to gravity and is opposite to the direction of motion, so \(a = -g = -9.8 \, \text{ms}^{-2}\).
- \(s\) is the displacement (maximum height, let's call it \(h\)).
Substituting the values:
\(0^2 = (10 \, \text{ms}^{-1})^2 + 2(-9.8 \, \text{ms}^{-2})h\)
\(0 = 100 - 19.6h\)
\(19.6h = 100\)
\(h = \frac{100}{19.6} \, \text{m} \approx 5.102 \, \text{m}\)
The displacement is upwards, with magnitude \(h\).
Step 4: Calculate Work Done by Gravity
The force of gravity \(F_g\) acts downwards. The displacement \(h\) is upwards. The angle \(\theta\) between the force of gravity and the displacement is \(180^\circ\). Therefore, \(\cos(180^\circ) = -1\).
Work done by gravity \(W_g = F_g \cdot h \cdot \cos(\theta)\)
\(W_g = (m \cdot g) \cdot h \cdot \cos(180^\circ)\)
\(W_g = (0.040 \, \text{kg} \times 9.8 \, \text{ms}^{-2}) \times \left(\frac{100}{19.6} \, \text{m}\right) \times (-1)\)
\(W_g = (0.392 \, \text{N}) \times \left(\frac{100}{19.6} \, \text{m}\right) \times (-1)\)
\(W_g = 0.392 \times \frac{100}{19.6} \times (-1)\)
Notice that \(0.392 = 0.040 \times 9.8\) and \(19.6 = 2 \times 9.8\). So, the calculation becomes:
\(W_g = (0.040 \times 9.8) \times \left(\frac{100}{2 \times 9.8}\right) \times (-1)\)
The term \(9.8\) cancels out:
\(W_g = 0.040 \times \frac{100}{2} \times (-1)\)
\(W_g = 0.040 \times 50 \times (-1)\)
\(W_g = 2.0 \times (-1)\)
\(W_g = -2.0 \, \text{J}\)
The work done by the force of gravity while the particle goes up is -2.0 J. The negative sign indicates that the force of gravity opposes the displacement.
| Quantity | Value |
|---|---|
| Mass (m) | \(0.040 \, \text{kg}\) |
| Initial Velocity (u) | \(10 \, \text{ms}^{-1}\) |
| Final Velocity (v) | \(0 \, \text{ms}^{-1}\) |
| Acceleration (a = -g) | \(-9.8 \, \text{ms}^{-2}\) |
| Displacement (h) | \(\approx 5.102 \, \text{m}\) |
| Force of Gravity (Fg = mg) | \(0.392 \, \text{N}\) |
| Angle between Fg and h | \(180^\circ\) |
| Work Done (Wg) | \(-2.0 \, \text{J}\) |
Thus, the work done by the force of gravity during the time the particle goes up is -2.0 J.
Revision Table: Key Concepts for Work Done by Gravity
| Concept | Description | Formula/Principle |
|---|---|---|
| Work Done (General) | Transfer of energy by a force causing displacement. | \(W = F \cdot d \cdot \cos(\theta)\) |
| Force of Gravity | Force exerted by Earth on an object towards its center. | \(F_g = m \cdot g\) |
| Work Done by Gravity (Upward Motion) | Gravity acts downwards, displacement is upwards. Work is negative. | \(W_g = -mgh\) (where h is vertical displacement upwards) |
| Work-Energy Theorem | The net work done on an object equals its change in kinetic energy. | \(W_{net} = \Delta KE\) |
Additional Information: Potential Energy and Work Done
The work done by the force of gravity is closely related to the concept of gravitational potential energy. The change in gravitational potential energy (\(\Delta PE\)) when an object moves from height \(h_1\) to \(h_2\) is given by \(\Delta PE = mg(h_2 - h_1)\).
The work done by the conservative force (like gravity) is equal to the negative of the change in potential energy associated with that force.
\(W_{\text{gravity}} = -\Delta PE_{\text{gravity}}\)
In this problem, the particle starts at some initial height (let's say \(y_i = 0\)) and reaches a final height \(y_f = h\). The change in potential energy is \(\Delta PE = mg(h - 0) = mgh\).
So, the work done by gravity is \(W_g = -\Delta PE = -mgh\). This matches our calculation using the force and displacement method. The negative sign indicates that gravity removes energy from the particle's kinetic energy as it rises, storing it as potential energy.
Understanding the relationship between work done by conservative forces and potential energy change is fundamental in physics and energy conservation principles.
Important Questions from Conservation of Mechanical Energy
A ball is thrown up at a speed of 2m/s. If g = 10m/s 2, then find the maximum height the ball will reach?
Find the work done by the force of gravity during the time a particle of mass 50 gm goes up on being thrown vertically upwards with a speed of 10 m/s.
A uniform chain of mass m and length l is placed on a smooth horizontal table such that \(\frac{1}{4}\)th of its length is hanging from the edge of the table. The chain slips down. Find the kinetic energy of the chain when half of its length is hanging from the edge of the table.
Which of the following equation is also a special case of the work-energy (WE) theorem? (where a is acceleration, u and v are the initial and final speeds and s the distance traversed.)
When a particle is projected upwards, its kinetic energy
