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An object is moving with a velocity of 5ms-1. At what height must a similar body...


Question

An object is moving with a velocity of 5ms-1. At what height must a similar body be situated to have a potential energy equal in value with the kinetic energy of the moving body? [g ≈ 10ms-2]

Options

A) 1.3m

B) 25.0m

C) 1.0m

D) 20.0m


The correct answer is A.

Explanation:

Let's start from K.E = P.E:

\(\text{mgh} = \frac{1}{2}\text{mv}^2\)

\(\text{mgh} = \cfrac{\text{mv}^2}{2}\)

Cross multiply:

It gives us:

\(2\text{mgh} = \text{mv}^2\)

Divide both sides by m:

It gives:

\(2\text{gh} = \text{v}^2\)

To get the value of h, divide both sides by 2g

The formula to use will become:

\(\text{h} = \cfrac{\text{v}^2}{2\text{g}}\)

Then \(\text{v} = 5\text{ms}^{-1}\)

g = 10. Put this into the above formula:

\(\text{h} = \cfrac{5^2}{2 \times 10}\)

\(\text{h} = \cfrac{25}{20}\)

\(\text{h} = 1.3\text{m}\)

Explanation provided by Idowu Taye Ebenezer


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Dicussion (1)

  • Let's start from K.E = P.E:

    \(\text{mgh} = \frac{1}{2}\text{mv}^2\)

    \(\text{mgh} = \cfrac{\text{mv}^2}{2}\)

    Cross multiply:

    It gives us:

    \(2\text{mgh} = \text{mv}^2\)

    Divide both sides by m:

    It gives:

    \(2\text{gh} = \text{v}^2\)

    To get the value of h, divide both sides by 2g

    The formula to use will become:

    \(\text{h} = \cfrac{\text{v}^2}{2\text{g}}\)

    Then \(\text{v} = 5\text{ms}^{-1}\)

    g = 10. Put this into the above formula:

    \(\text{h} = \cfrac{5^2}{2 \times 10}\)

    \(\text{h} = \cfrac{25}{20}\)

    \(\text{h} = 1.3\text{m}\)