(B) Given: Mass $m = 15 \,kg$, initial velocity $u = 0$. At $t = 3 \,s$, power $P = 5 \,W$. Since $P = F \cdot v$ and $v = at = (F/m)t$, we have $P =

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(B) Given: Mass $m = 15 \,kg$, initial velocity $u = 0$. At $t = 3 \,s$, power $P = 5 \,W$. Since $P = F \cdot v$ and $v = at = (F/m)t$, we have $P = F \cdot (F/m)t = (F^2/m)t$. Substituting the values: $5 = (F^2 / 15) \times 3$. $F^2 = (5 \times 15) / 3 = 25$, so $F = 5 \,N$. Now, at $t = 4 \,s$, the velocity $v' = (F/m)t = (5/15) \times 4 = 4/3 \,m/s$. The instantaneous power $P' = F \cdot v' = 5 \times (4/3) = 20/3 \,W \approx 6.67 \,W$.

Class 11 Physics Work, Energy, Power and Collision Power

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$A$ force acts on a body of mass $15 \,kg$, initially at rest. If the instantaneous power due to the force at the end of the third second is $5 \,W$, then the instantaneous power (in $W$) at the end of the $4 \,s$ will be

TS EAMCET 2021 All TS EAMCET

A
$6.33$
B
$6.67$
C
$6.29$
D
$6.94$

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Work, Energy, Power and Collision

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$A$ force acts on a body of mass $15 \,kg$, initially at rest. If the instantaneous power due to the force at the end of the third second is $5 \,W$, then the instantaneous power (in $W$) at the end of the $4 \,s$ will be

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Water falls down a $500.0 \,m$ shaft to reach a turbine which generates electricity. How many $m^3$ of water must fall per second in order to generate $1.00 \times 10^9 \,W$ of power? (Assume $50 \%$ efficiency of conversion and $g=10 \,ms^{-2}$)

$A$ train of mass $2.05 \times 10^6 \ kg$ changes its velocity from $5 \ m/s$ to $25 \ m/s$ in $5 \ minutes$. The power of the engine is . . . . . . $MW$.

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