Question

An ice block on a frictionless path starts at rest and is released on the path. How how much kinetic energy does it have at point A?

Answer 1

Answer:

Kinetic energy = 44.1m Joules

where m represents the mass of the ice block

Explanations:

The Kinetic Energy is given by the formula:

$KE\text{ = }\frac{1}{2}mv^2$

Since the path is frictionless, energy at the initial point equals the energy at point A

The energy at the initial point is:

PE = mgh

h = 4.5 m, g = 9.8 m/s²

PE = m x 9.8 x 4.5

PE = 44.1m

The energy at the point A:

$KE\text{ = }\frac{1}{2}mv^2$

Since PE = KE

$\begin{gathered} 44.1m\text{ = }\frac{1}{2}mv^2 \\ 88.2m=mv^2 \\ v^2\text{ = }\frac{88.2m}{m} \\ v\text{ = }\sqrt[]{88.2} \\ v\text{ = }9.4\text{ m/s} \end{gathered}$

The KE energy will therefore be calculated as:

$\begin{gathered} KE\text{ = }\frac{1}{2}mv^2 \\ KE\text{ = 0.5}\times\text{ m }\times9.4^2 \\ KE\text{ = 44.1m } \end{gathered}$

Kinetic energy = 44.1m Joules