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प्रश्न
A car weighing 1400 kg is moving at a speed of 54 km/h up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10 m above the point, calculate the work done against friction (negative of the work done by friction).
उत्तर
\[\text{ Given }, \]
\[\text{ Mass of the car, m = 1400 kg }\]
\[\text{ h = 10 m }\]
\[\text{ Since the car is moving when the motor stops, it has kinetic energy . Thus } \]
\[ \text{ v}_i = 54 \text{ km/h } \times \frac{5}{18} = 15 \text{ m/s }\]
\[ \text{v}_\text{f} = 0\]
\[ \bigtriangleup \text{K} = \frac{1}{2} {{\text{ mv }}_\text{f}}^2 - \frac{1}{2} {{\text{mv}}_\text{i}}^2 \]
\[ \bigtriangleup \text{K} = \frac{1}{2} \times 1400\left( 0^2 - {15}^2 \right)\]
\[ \bigtriangleup \text{ K = - 157500 J }\]
\[\text{ Let the gravitational potential energy be zero at the starting point } . \]
\[\text{ Then the potential energy at the terminal is } \]
\[ \text{U}_\text{i} = 0\]
\[ \text{U}_\text{f} = \text{mgh}\]
\[ \text{U}_\text{f} = 1400 \times 9 . 8 \times 10 = 137200 \text{ J }\]
\[ \bigtriangleup \text{ U = U}_\text{f} - \text{U}_\text{i} \]
\[ \bigtriangleup \text{ U = 137200 - 0 = 137200 J }\]
\[\text{ Let W be the work done against friction during ascent .} \]
\[\text{Then " - W " is the work done by the frictional force }. \]
\[ - \text{W} = \bigtriangleup \text{K }+ \bigtriangleup \text{U}\]
\[ - \text{W} = - 157500 + 137200\]
\[ - \text{W = - 20300 J}\]
\[\text{W = 20300 J}\]
So, work done against friction is 20,300 joule.
