Answer:
The value is [tex]P =3.294 \ W[/tex]
Explanation:
From the question we are that
The velocity [tex]v = 0.540 \ m/s[/tex]
The time taken is [tex]t = 27.0 ms = 27.0 *10^{-3} \ s[/tex]
The total mass of the train is [tex]m = 610 \ g = 0.610 \ kg[/tex]
Generally the average power delivered is mathematically represented as
[tex]P =\frac{KE }{t}[/tex]
[tex]P =\frac{ \frac{1}{2} * m * v^2 }{t}[/tex]
=> [tex]P =\frac{ \frac{1}{2} * 0.610 * 0.540 ^2 }{ 27.0 *10^{-3}}[/tex]
=> [tex]P =3.294 \ W[/tex]
Answer:
3.29 Watts
Explanation:
Given that
Initial velocity, u = 0 m/s
Final velocity, v = 0.54 m/s
Change in time, Δt = 27 ms
Mass of the train, m = 610 g = 0.61 kg
Power of any object is given as
P = Work done / time
P = W / t
Work done in this question, W is
W = 1/2mv²
W = 1/2 * 0.61 * 0.54²
W = 305 * 0.2916
W = 0.0889 kgm²/s²
Plugging this value into the equation we'd stated earlier, we have
P = W / t
P = 0.0889 / 27*10^-3
P = 3.293 Watts
Therefore, the average power delivered to the train during its acceleration is 3.29 Watts
