Question 240: Work, Energy and Power & Circular Motion

Question

A particle of mass $$10 \mathrm{~g}$$ moves in a straight line with retardation $$2 x$$, where $$x$$ is the displacement in SI units. Its loss of kinetic energy for above displacement is $$\left(\frac{10}{x}\right)^{-n}$$ J. The value of $$\mathrm{n}$$ will be __________

Accepted Answer

The work done against the retarding force is indeed equal to the loss in kinetic energy.
The force acting on the particle due to retardation is given by $F = ma = -2mx$.
When we integrate this force over the displacement from $0$ to $x$, we get:
$$\Delta KE = W = \int F \cdot dx = \int (-2mx) \, dx = -mx^2$$
The negative sign indicates that this is a loss of kinetic energy.
The problem states that the loss in kinetic energy is also given by $\left(\frac{10}{x}ight)^{-n}$ J. Therefore, we have:
$$-mx^2 = \left(\frac{10}{x}ight)^{-n}$$
Because this is a loss of kinetic energy, we should consider the absolute value. Hence,
$$mx^2 = \left(\frac{10}{x}ight)^{-n}$$
Substituting the given mass $m = 10 \, 	ext{g} = 0.01 \, 	ext{kg}$, we get:
$$0.01x^2 = \left(\frac{10}{x}ight)^{-n}$$
This simplifies to:
$$x^2 = \left(\frac{10}{x}ight)^{-n}$$
Comparing the two sides, we can see that $n = 2$.

Work, Energy and Power & Circular Motion

A particle of mass $10 \mathrm{~g} moves in a straight line with retardation 2 x, where x is the displacement in SI units. Its loss of kinetic energy for above displacement is \left(\frac{10}{x}\right)^{-n} J. The value of \mathrm{n}$ will be __________

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