If the resultant force acting on a particle is perpendicular to its velocity then- 1. The speed of the particle changes. 2. The speed of the particle does not change. 3. Kinetic energy of the particle does not change. Choose the correct code.
Only 2 and 3
When a resultant force acts on a particle, its effect on the particle's motion depends on the direction of the force relative to the particle's velocity. A particularly important case is when the resultant force is perpendicular to the particle's instantaneous velocity.
The work done by a force (\(\vec{F}\)) during a small displacement (\(d\vec{r}\)) is given by the dot product \(dW = \vec{F} \cdot d\vec{r}\). The instantaneous displacement \(d\vec{r}\) is always in the direction of the instantaneous velocity (\(\vec{v}\)). Therefore, if the force \(\vec{F}\) is perpendicular to the velocity \(\vec{v}\), it is also perpendicular to the displacement \(d\vec{r}\).
In this case, the angle between \(\vec{F}\) and \(d\vec{r}\) is 90 degrees. The work done is:
\[ dW = |\vec{F}| |d\vec{r}| \cos(90^\circ) = |\vec{F}| |d\vec{r}| \times 0 = 0 \]The total work done by the resultant force over any path will be the sum of these infinitesimal work elements, which is also zero.
\[ W = \int dW = \int 0 = 0 \]The work-energy theorem states that the change in kinetic energy (\(\Delta KE\)) of a particle is equal to the total work done (\(W\)) on it by the resultant force.
\[ \Delta KE = W \]Since the work done by the resultant force is zero when it is perpendicular to the velocity, the change in kinetic energy is also zero:
\[ \Delta KE = 0 \]This means that the kinetic energy of the particle remains constant.
\[ KE = \text{constant} \]Kinetic energy is defined as \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the particle and \(v\) is its speed. For a given particle, the mass \(m\) is constant. If the kinetic energy \(KE\) is constant, then \(\frac{1}{2}mv^2\) must be constant. Since \(\frac{1}{2}m\) is a non-zero constant, \(v^2\) must also be constant.
\[ v^2 = \frac{2KE}{m} = \text{constant} \]If \(v^2\) is constant, then the speed \(v\) must also be constant (assuming \(v \ge 0\)).
Let's examine each statement based on our understanding:
As we derived, if the resultant force is perpendicular to the velocity, the speed of the particle remains constant. Therefore, this statement is incorrect.
Our analysis shows that the speed remains constant under these conditions. Therefore, this statement is correct.
Since the work done by the perpendicular force is zero, the change in kinetic energy is zero, meaning the kinetic energy remains constant. Therefore, this statement is correct.
Based on the analysis, statements 2 and 3 are correct.
| Statement | Analysis | Correct/Incorrect |
|---|---|---|
| 1. The speed of the particle changes. | Work done by force \(\perp\) velocity is zero. Zero work means constant KE. Constant KE means constant speed. | Incorrect |
| 2. The speed of the particle does not change. | Constant KE implies constant speed. | Correct |
| 3. Kinetic energy of the particle does not change. | Work done by force \(\perp\) velocity is zero. Work-energy theorem says \(\Delta KE = W\). So \(\Delta KE = 0\). | Correct |
When the resultant force on a particle is always perpendicular to its velocity, the force only changes the direction of the velocity, not its magnitude (speed). This type of force is often called a centripetal force, which causes the particle to move along a curved path, such as circular motion.
| Condition | Work Done | Change in KE | Change in Speed | Effect on Motion |
|---|---|---|---|---|
| Force \(\perp\) Velocity | Zero (\(W=0\)) | Zero (\(\Delta KE=0\)) | Zero (\(\Delta v=0\)) | Changes direction only (e.g., circular motion) |
| Force || Velocity (same direction) | Positive | Positive | Positive (speed increases) | Changes magnitude and direction (if initial motion is not in line) |
| Force || Velocity (opposite direction) | Negative | Negative | Negative (speed decreases) | Changes magnitude and direction (if initial motion is not in line) |
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