Work done due to friction — task. Physics CBSE Live product, Class 9 (2025-26).

Calculate the correct value:

A girl having mass of \(40\ kg\) sits on a trolley of mass \(6\ kg\). The trolley is given an initial velocity of \(5\ m s^{-1}\) by applying a force. The trolley comes to rest after traversing a distance of \(13\ m\).

a. How much work is done on the trolley?

b. How much work is done by the girl?

Equation of motion,

i^{i} - i^{i} = i \times i \times i

Force is calculated by:

\(F\ =\ \frac{m}{s}\)
\(F\ =\ m \times a\)
\(F\ =\ \frac{m}{a}\)
\(F\ =\ m \times s\)

The formula to calculate the Work done(\(W\)) is,

\(W\ =\ F \times g\)
\(W\ =\ \frac{F}{g}\)
\(W\ =\ F \times s\)
\(W\ =\ \frac{F}{s}\)

a. Work done on the trolley \(=\) \(J\)

b. Work done by the girl \(=\) \(J\)

[Note: Enter the answer upto 2 decimal places.]

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9. Work done due to friction

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