1. Tardigrade
  2. Question
  3. Physics
  4. A car weighing 1400 kg is moving at a speed of 54 kmh -1 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, then the work done against friction (negative of the work done by the friction) is (Take g=10 ms -2 )

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Q. A car weighing $1400\, kg$ is moving at a speed of $54\, kmh ^{-1}$ 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, then the work done against friction (negative of the work done by the friction) is (Take $g=10\, ms ^{-2}$ )

Work, Energy and Power

Solution:

Here, $1400 \times 10 \times 10+W=\frac{1}{2} \times 15 \times 15$
or $W=700 \times 15 \times 15-1400 \times 10 \times 10$
$=700(225-200) J$
$=700 \times 25\, J =17.5\, kJ$