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.
-2.5 J
Understanding Work Done by Gravity During Vertical Motion
Let's analyze the question about the work done by the force of gravity on a particle moving vertically upwards. The force of gravity always acts downwards. When a particle is thrown vertically upwards, its displacement is in the upward direction. The work done by a force is defined as the dot product of the force and the displacement vectors, which is given by $W = |\vec{F}| |\vec{d}| \cos(\theta)$, where $\theta$ is the angle between the force and displacement vectors.
In this case:
- Force ($\vec{F}$) is the force of gravity, acting downwards.
- Displacement ($\vec{d}$) is upwards as the particle goes up.
- The angle ($\theta$) between the downward force and the upward displacement is $180^\circ$.
- The force of gravity is $F_g = mg$, where $m$ is the mass and $g$ is the acceleration due to gravity.
Since $\cos(180^\circ) = -1$, the work done by gravity during upward motion is always negative.
Given Information:
- Mass of the particle, $m = 50 \text{ gm} = 50 \times 10^{-3} \text{ kg} = 0.05 \text{ kg}$
- Initial velocity, $v_0 = 10 \text{ m/s}$ upwards
- We will take the acceleration due to gravity, $g = 10 \text{ m/s}^2$ (as this yields one of the options).
Method 1: Calculating Work Using Force and Displacement (Maximum Height)
First, let's find the maximum height the particle reaches. At the maximum height, the final velocity ($v$) is $0 \text{ m/s}$. We can use the kinematic equation:
\(v^2 = v_0^2 + 2ah\)
Here, $v=0$, $v_0=10 \text{ m/s}$, and acceleration $a$ due to gravity acting downwards is $a = -g = -10 \text{ m/s}^2$ (negative sign because it opposes upward motion).
\(0^2 = (10 \text{ m/s})^2 + 2(-10 \text{ m/s}^2)h\)
\(0 = 100 - 20h\)
\(20h = 100\)
\(h = \frac{100}{20} \text{ m}\)
\(h = 5 \text{ m}\)
The maximum height reached is 5 m.
Now, calculate the force of gravity acting on the particle:
\(F_g = mg\)
\(F_g = (0.05 \text{ kg}) \times (10 \text{ m/s}^2)\)
\(F_g = 0.5 \text{ N}\)
The displacement is upwards ($h=5 \text{ m}$), and the force of gravity is downwards ($F_g=0.5 \text{ N}$). The angle between them is $180^\circ$. The work done by gravity is:
\(W_g = F_g \times h \times \cos(180^\circ)\)
\(W_g = (0.5 \text{ N}) \times (5 \text{ m}) \times (-1)\)
\(W_g = -2.5 \text{ J}\)
Method 2: Using the Work-Energy Theorem
The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy. In this scenario, if we only consider gravity (ignoring air resistance), the work done by gravity is the net work done on the particle during its upward motion until it stops at the maximum height.
Initial kinetic energy ($KE_i$):
\(KE_i = \frac{1}{2}mv_0^2\)
\(KE_i = \frac{1}{2}(0.05 \text{ kg})(10 \text{ m/s})^2\)
\(KE_i = \frac{1}{2}(0.05)(100) \text{ J}\)
\(KE_i = 0.025 \times 100 \text{ J}\)
\(KE_i = 2.5 \text{ J}\)
Final kinetic energy ($KE_f$) at the maximum height (where $v=0$):
\(KE_f = \frac{1}{2}mv^2\)
\(KE_f = \frac{1}{2}(0.05 \text{ kg})(0 \text{ m/s})^2\)
\(KE_f = 0 \text{ J}\)
Change in kinetic energy ($\Delta KE$):
\(\Delta KE = KE_f - KE_i\)
\(\Delta KE = 0 \text{ J} - 2.5 \text{ J}\)
\(\Delta KE = -2.5 \text{ J}\)
According to the Work-Energy Theorem, the work done by gravity ($W_g$) is equal to the change in kinetic energy:
\(W_g = \Delta KE\)
\(W_g = -2.5 \text{ J}\)
Both methods yield the same result. The work done by the force of gravity is -2.5 J.
| Quantity | Symbol | Value | Units |
|---|---|---|---|
| Mass | m | 50 gm = 0.05 | kg |
| Initial Velocity | v<sub>0</sub> | 10 | m/s |
| Acceleration due to gravity | g | 10 | m/s<sup>2</sup> |
| Maximum Height | h | 5 | m |
| Force of Gravity | F<sub>g</sub> | 0.5 | N |
| Work Done by Gravity | W<sub>g</sub> | -2.5 | J |
Revision Table: Key Concepts
| Concept | Definition/Formula | Notes for Vertical Motion |
|---|---|---|
| Work Done by a Force | $W = |\vec{F}| |\vec{d}| \cos(\theta)$ | Negative if force opposes displacement (like gravity on upward motion) |
| Force of Gravity | $F_g = mg$ | Acts downwards |
| Work-Energy Theorem | $W_{net} = \Delta KE = KE_f - KE_i$ | If gravity is the only force doing work, $W_g = \Delta KE$ |
| Kinematic Equation | $v^2 = v_0^2 + 2ah$ | Useful for finding displacement ($h$) when initial/final velocity is known |
Additional Information: Understanding Work Done by Gravity
Work done by gravity depends on the vertical displacement. If an object moves downwards, gravity does positive work. If it moves upwards, gravity does negative work. If it moves horizontally, gravity does zero work. The total work done by gravity over a complete round trip (up and down) is zero because the net vertical displacement is zero.
The work done by gravity is related to the change in gravitational potential energy ($\Delta PE_g$). Specifically, $W_g = -\Delta PE_g$. As the particle moves up, its potential energy increases ($\Delta PE_g > 0$), so the work done by gravity is negative ($W_g < 0$). As it moves down, potential energy decreases ($\Delta PE_g < 0$), and work done by gravity is positive ($W_g > 0$).
In this problem, the particle moves to a higher point, increasing its potential energy, which is why the work done by gravity is negative.
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?
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.
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
