A force applied by an engine of a train of mass 2.05 × 106 kg changes its velocity from 5 m / s to 35 m / s in 15 minutes. The power of the engine is

Question

A force applied by an engine of a train of mass 2.05 × 106 kg changes its velocity from 5 m / s to 35 m / s in 15 minutes. The power of the engine is (A) 1.36 hp (B) 1.36 MW (C) 1.36 W (D) None

Tardigrade Admin · Accepted Answer

P=(w/t)=(Δ K/t)=(1/2) × (m(v2-u2)/t) =(2.05 × 106((35)2-(5)2)/2 × 15 × 60) P=(2.05 × 106(1225-25)/1800) =1.36 × 106=1.36 M Watt

Q. A force applied by an engine of a train of mass $2.05 \times 1 0^{6} k g$ changes its velocity from $5 m / s$ to $35 m / s$ in $15$ minutes. The power of the engine is

$1.36 h p$

$1.36 M W$

$1.36 W$

None

$P = \frac{w}{t} = \frac{Δ K}{t} = \frac{1}{2} \times \frac{m ( v ^{2} - u ^{2} )}{t}$
$= \frac{2.05 \times 1 0 ^{6} ( ( 35 ) ^{2} - ( 5 ) ^{2} )}{2 \times 15 \times 60}$
$P = \frac{2.05 \times 1 0 ^{6} ( 1225 - 25 )}{1800}$
$= 1.36 \times 1 0^{6} = 1.36 M$ Watt

Q. A force applied by an engine of a train of mass 2.05×106kg changes its velocity from 5m/s to 35m/s in 15 minutes. The power of the engine is

1.36hp

1.36MW

1.36W