Q.A particle of mass $1\ \text{kg}$, initially at rest at $x = 0$, moves along the $x$-axis under a force $F(x) = (10 - 2x)\ \text{N}$ (with $x$ in metres). Using the work-energy theorem, find the speed of the particle when it reaches $x = 5\ \text{m}$.

Given: mass $m=1\ \text{kg}$, force $F(x) = (10-2x)\ \text{N}$, starting from rest at $x=0$, find speed at $x=5\ \text{m}$.

Work done as the particle moves from $x=0$ to $x=5\ \text{m}$: $W = \int_0^5 (10-2x)\,dx = \Big[10x - x^2\Big]_0^5 = (50 - 25) - (0-0) = 25\ \text{J}$

By the work-energy theorem, since the particle starts from rest ($K_i=0$): $W = \Delta K = \frac{1}{2}mv^2 \quad\Longrightarrow\quad 25 = \frac{1}{2}(1)v^2 = \frac{v^2}{2}$ $v^2 = 50 \quad\Longrightarrow\quad v = \sqrt{50} = 5\sqrt{2} \approx 7.07\ \text{m/s}$

(Note: the force $F(x)=10-2x$ becomes zero exactly at $x=5\ \text{m}$ and would turn negative beyond that point, but this does not affect the answer here, since only the net work done up to $x=5\ \text{m}$ -- where the force has not yet turned negative -- is required.)

Integrate the given $F(x)$ from $0$ to $5\ \text{m}$ to find the work done, then apply the work-energy theorem with $u=0$ to solve for the final speed.

• Forgetting the negative sign in $-2x$ when integrating, giving $10x+x^2$ instead of $10x-x^2$.
• Not evaluating the definite integral correctly at both limits, or forgetting to subtract the value at $x=0$ (which is zero here, but should still be checked explicitly).