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 )

Question

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 ) (A) 10 kJ (B) 15 kJ (C) 17.5 kJ (D) 25 kJ

Tardigrade Admin · Accepted Answer

Here, 1400 × 10 × 10+W=(1/2) × 15 × 15 or W=700 × 15 × 15-1400 × 10 × 10 =700(225-200) J =700 × 25 J =17.5 kJ

Q. A car weighing 1400kg is moving at a speed of 54kmh−1 up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10m above the point, then the work done against friction (negative of the work done by the friction) is (Take g=10ms−2 )

10 kJ

15 kJ

17.5 kJ

25 kJ

Here, 1400×10×10+W=21​×15×15 or W=700×15×15−1400×10×10 =700(225−200)J =700×25J=17.5kJ

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