Question

A body of mass 20 kg is dropped from a height of 100m .Find its Kinetic energyand potential energy after(i) one second (ii) Two second.

Answer 1

Given

m = 20 kg

h = 100m

Kinetic energy and potential energy after

(i) 1 sec

(ii) 2 sec

Procedure

The motion we are analyzing is a free fall, therefore the acceleration faced by the object is 9.8m/s^2 which is gravity.

Kinetic energy is a form of energy that an object or a particle has by reason of its motion.

Let's first calculate the velocity after 1s

`\begin{gathered} v^{}_f=v^{}_o-gt \\ v_f=-9.8m/s^2\cdot1s \\ v_f=-9.8m/s \end{gathered}`

Now with the velocity, we can calculate the kinetic energy

`\begin{gathered} K=(1)/(2)mv^2 \\ K=(1)/(2)20kg\cdot(9.8m/s)^2 \\ K=960.4\text{ J} \end{gathered}`

Now let's calculate the potential energy after 1s. Let's first calculate the displacement after that time

`\begin{gathered} y=v_ot-(1)/(2)gt^2 \\ y=-(1)/(2)gt^2 \\ y=-(1)/(2)\cdot9.8m/s^2\cdot(1s)^2 \\ y=4.9\text{ m} \end{gathered}`

Now let us proceed to calculate the potential energy

`\begin{gathered} U=\text{mgh} \\ U=20kg\cdot9.8m/s^2\cdot(100m-4.9m) \\ U=18639.6\text{ J} \end{gathered}`

For 1s:

K = 960.4 J

U = 18639.6 J

In a very similar way we are going to calculate for the second 2

Let's first calculate the velocity after 2s

`\begin{gathered} v_f=-9.8m/s^2\cdot2s \\ v_f=-19.6m/s \end{gathered}`

Now with the velocity, we can calculate the kinetic energy

`\begin{gathered} K=(1)/(2)mv^2 \\ K=(1)/(2)20kg\cdot(19.6m/s)^2 \\ K=3841.6\text{ J} \end{gathered}`

Now let's calculate the potential energy after 2s. Let's first calculate the displacement after that time

`\begin{gathered} y=v_ot-(1)/(2)gt^2 \\ y=-(1)/(2)gt^2 \\ y=-(1)/(2)\cdot9.8m/s^2\cdot(2s)^2 \\ y=19.6\text{ m} \end{gathered}`

Now let us proceed to calculate the potential energy

`\begin{gathered} U=\text{mgh} \\ U=20kg\cdot9.8m/s^2\cdot(100m-19.6m) \\ U=15758.4\text{ J} \end{gathered}`