A force applied by an engine of a train of mass 2.05× 106kg changes its velocity from 5ms- 1 to 25ms- 1 in 5 minutes. The power of the engine is :-

Question

A force applied by an engine of a train of mass 2.05× 106kg changes its velocity from 5ms- 1 to 25ms- 1 in 5 minutes. The power of the engine is :- (A) 1.025MW (B) 2.05MW (C) 5MW (D) 6MW

Tardigrade Admin · Accepted Answer

Power =(Work done / time )=((1/2) m (v2 - u2)/t) =(1/2)× (2 . 05 × (10)6 × ((25)2 - 52)/5 × 60) P=2.05× 106Watt=2.05MW

Q. A force applied by an engine of a train of mass $2.05 \times 1 0^{6} k g$ changes its velocity from $5 m s^{- 1}$ to $25 m s^{- 1}$ in $5$ minutes. The power of the engine is :-

$1.025 M W$

$2.05 M W$

$5 M W$

$6 M W$

Power $= \frac{W or k d o n e}{t im e} = \frac{\frac{1}{2} m ( v ^{2} - u ^{2} )}{t}$
$= \frac{1}{2} \times \frac{2.05 \times ( 10 ) ^{6} \times ( ( 25 ) ^{2} - 5 ^{2} )}{5 \times 60}$
$P = 2.05 \times 1 0^{6} Wa tt = 2.05 M W$

Q. A force applied by an engine of a train of mass 2.05×106kg changes its velocity from 5ms−1 to 25ms−1 in 5 minutes. The power of the engine is :-

1.025MW

2.05MW

5MW

6MW