Question

A 8.1 kg object initially at rest is pushed down a 15.0 m tall hill. What is the speed of the object at the bottom of the hill?

Answer 1

Answer: 17.14 m/s

Explanation:

We can solve this problem applying the Conservation of Mechanical energy (which is the sum of the kinetic energy $K$ and potential energy $P$), where the total initial mechanical energy $E_{o}$ must be equal to the total final mechanical energy $E_{f}$:

$E_{o}=E_{f}$ (1)

Being:

$E_{o}=K_{o}+P_{o}=\frac{1}{2}mV_{o}^{2}+mgh_{o}$ (2)

$E_{f}=K_{f}+P_{f}=\frac{1}{2}mV_{f}^{2}+mgh_{f}$ (3)

Where:

$m=8.1 kg$ is the mass of the object

$V_{o}=0 m/s$ is the initial velocity (the object was at rest)

$g=9.8 m/s^{2}$ is the acceleration due gravity

$h_{o}=15 m$ is the object's initial height or position

$V_{f}$ is the final velocity

$h_{f}=0 m$ is the object's final height or position

Solving and applying the given conditions:

$\frac{1}{2}mV_{o}^{2}+mgh_{o}=\frac{1}{2}mV_{f}^{2}+mgh_{f}$ (4)

$mgh_{o}=\frac{1}{2}mV_{f}^{2}$ (5)

Finding $V_{f}$:

$V_{f}=\sqrt{2gh_{o}}$ (6)

$V_{f}=\sqrt{2(9.8 m/s^{2})(15 m)}$ (7)

Finally:

$V_{f}=17.14 m/s$