A.3 Work, energy and power

The child starts from rest at height $h=8.0$ m. Since the slide is frictionless, mechanical energy is conserved: all gravitational potential energy converts to kinetic energy.

Setting $E_p$ at the top equal to $E_k$ at the bottom (taking the pool as the reference level): $mgh=\frac{1}{2}m{v}^2$

Mass cancels from both sides: $v=\sqrt{2gh}=\sqrt{2\times 9.81\times 8.0}$

$v=\sqrt{156.96}\approx 12.5{\text{ m s}}^{-1}$ The child's speed entering the pool is 12.5 m s${}^{-1}$.

This question requires applying conservation of mechanical energy on a frictionless surface, using $mgh=\frac{1}{2}m{v}^2$ to find speed.

9.81 m s${}^{-1}$ equals $g$, which is not the correct formula result. This is a common distractor involving misuse of $g$.

17.6 m s${}^{-1}$ would result from using $v=2gh$ without the square root, which is an algebraic error.

15.7 m s${}^{-1}$ results from incorrectly computing $\sqrt{2\times 9.81\times 8.0}$ as $\sqrt{245}$, possibly from using $g=9.81$ with an arithmetic error. The correct value is $\sqrt{156.96}\approx 12.5$.

12.5 m s${}^{-1}$ is correct, obtained from $v=\sqrt{2\times 9.81\times 8.0}\approx 12.5$ m s${}^{-1}$.

All elastic potential energy in the compressed spring converts to kinetic energy of the car: $E_e=E_k$ $\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2$

$E_e=\frac{1}{2}\times 500\times (0.10{)}^2=\frac{1}{2}\times 500\times 0.01=2.5\text{ J}$

$\frac{1}{2}m{v}^2=2.5$ $\frac{1}{2}\times 0.20\times v^2=2.5$ $v^2=\frac{2.5}{0.10}=25$

$v=\sqrt{25}=5.0$ Wait — let me recheck: $v^2=\frac{2\times 2.5}{0.20}=\frac{5.0}{0.20}=25$, so... Actually $v=\sqrt{25}$... but that gives 5.0. Let me recalculate: $v^2=\frac{2\times 2.5}{0.20}=25$, $v=5.0$ m s${}^{-1}$. Hmm — wait. $k=500$, $x=0.10$: $E_e=\frac{1}{2}(500)(0.01)=2.5$ J; $v=\sqrt{\frac{2\times 2.5}{0.20}}=\sqrt{25}=5.0$ m s${}^{-1}$.

We need $E_e=\frac{1}{2}k{x}^2=\frac{1}{2}(500)(0.10{)}^2=2.5$ J, then $v=\sqrt{\frac{2E_e}{m}}=\sqrt{\frac{5.0}{0.20}}=\sqrt{25}=5.0$ m s${}^{-1}$.

0.50 m s${}^{-1}$ is far too small and would correspond to an error where $x$ or $k$ values are incorrectly used.

1.58 m s${}^{-1}$ would result from forgetting to square $x$, using $E_e=\frac{1}{2}kx$ rather than $\frac{1}{2}k{x}^2$.

2.50 m s${}^{-1}$ arises from an error such as $v=\sqrt{E_e/m}$ instead of $v=\sqrt{2E_e/m}$, omitting the factor of 2.

5.00 m s${}^{-1}$ is correct: $v=\sqrt{\frac{2\times 2.5}{0.20}}=\sqrt{25}=5.00$ m s${}^{-1}$.