A particle of mass 10 g moves in a straight line with retardation 2x, where x is the displacement in SI units. Its loss of kinetic energy for above displacement is (10 x)
Understand the Problem
The question involves a particle with a given mass and a specific retardation that varies with the displacement. We need to calculate the loss of kinetic energy as the particle moves a distance 'x'. This requires applying concepts of physics related to motion, forces, and energy.
Answer
The loss of kinetic energy as the particle moves a distance $x$ is given by $\Delta K.E. = \int_0^x f(x') \, dx'$.
Answer for screen readers
The loss of kinetic energy as the particle moves a distance $x$ is equal to the work done against the retardation, which can be expressed mathematically as:
$$ \Delta K.E. = W = \int_0^x f(x') , dx' $$
Steps to Solve
- Identify the initial kinetic energy
The initial kinetic energy ($K.E._i$) of a particle with mass $m$ and initial velocity $v$ is given by the formula:
$$ K.E._i = \frac{1}{2} mv^2 $$
- Determine the force acting on the particle
Given the retardation (deceleration) is described as a function of displacement, we express it as $a = -f(x)$, where $f(x)$ represents the function describing the retardation.
- Calculate the work done against the retardation
The work done ($W$) as the particle moves a distance $x$ can be calculated as:
$$ W = \int_0^x f(x') , dx' $$
- Calculate the final kinetic energy
Using the work-energy principle, the final kinetic energy ($K.E._f$) after traveling the distance $x$ is:
$$ K.E._f = K.E._i - W $$
- Calculate the loss of kinetic energy
The loss of kinetic energy ($\Delta K.E.$) is then given by:
$$ \Delta K.E. = K.E._i - K.E._f = W $$
This is the final expression we use to determine the loss of kinetic energy.
The loss of kinetic energy as the particle moves a distance $x$ is equal to the work done against the retardation, which can be expressed mathematically as:
$$ \Delta K.E. = W = \int_0^x f(x') , dx' $$
More Information
The loss of kinetic energy varies depending on how the retardation function $f(x)$ is defined. For constant retardation, this simplifies the calculation. If the function changes with displacement, integration helps capture the varying forces acting on the particle.
Tips
- Forgetting to account for the direction of retardation, which should be negative in relation to acceleration.
- Not using proper limits of integration when calculating work done over a given distance.
AI-generated content may contain errors. Please verify critical information
