Question

A 5 kg object is dropped from rest at the top of a 110 meter cliff. After it has fallen a distance of 10 meters, how much kinetic energy does the object have? [Use g = 10 m/s2]

Answer 1

Given data

*The given mass of the object is m = 5 kg

*The given height of the cliff is h = 110 m

*The given distance is s = 10 m

*The value of the acceleration due to gravity is g = 10 m/s^2

*The initial velocity of the object is u = 0 m/s

The formula for the final velocity of the object is given by the equation of motion as

`v^2=u^2+2gs`

Substitute the known values in the above expression as

`\begin{gathered} v^2=(0)^2+2(10)(10) \\ v=\sqrt[]{200} \\ =14.14\text{ m/s} \end{gathered}`

The formula for the kinetic energy of the object is given as

`U_k=(1)/(2)mv^2`

Substitute the known values in the above expression as

`\begin{gathered} U_k=(1)/(2)(5)(14.14)^2 \\ =499.84\text{ J} \\ \approx500\text{ J} \end{gathered}`

Hence, the kinetic energy of the object is U_k = 500 J