Powerful Block Push

Given that the applied force $F(t)$ is such that the power $P$ is a constant. Denote that mass and the instantaneous velocity of the be $m=\text{1500 kg}$ and $v(t)$ respectively.

$\begin{aligned} P & = {\color{#3D99F6}F(t)}v(t) & \small \color{#3D99F6} \text{Newton's 2nd law: } F = ma = m\frac {dv}{dt} \\ P & = {\color{#3D99F6}m}v\color{#3D99F6}\frac {dv}{dt} \\ \int_0^{10} P\ dt & = \int_0^{\frac {100 \times 10^3}{60\times 60}} 1500 v \ dv \\ Pt \ \bigg|_0^{10} & = \frac {1500 v^2}2 \ \bigg|_0^{\frac {100 \times 10^3}{60\times 60}} \\ \implies P & = \frac {1500}{2 \times 10}\left(\frac {100 \times 10^3}{60\times 60}\right)^2 \\ & \approx 57870.370 \text{ kW} \times \frac 1{\text{746 kW}} \\ & \approx \boxed {77.574} \text{ horsepower} \end{aligned}$

Velocity of body after $10s=100kph=\frac{250}{9} m/s=v$

$P=\frac{E}{t}$

$E_{kinetic}=\frac{1}{2}mv^2$

$=\frac{1500×250×250}{2×9×9}$

$=\frac{150×250×250}{3×9}$

$t=10s(given)$

So, $P_{horsepower}=\frac{E}{t}=\frac{150×250×250}{3×9×10×746}=\boxed{\boxed{77.5742..}}$

Note : $P$ is power, $E$ is energy, $t$ is time, $v$ is velocity , $m$ is mass

Since power is constant, work done by the force is proportional to time. By the Work-Energy theorem, work done by the force equals the change in kinetic energy of the body. Therefore, power is the ratio of the change in kinetic energy of the body and time. Since initial K.E. of the body is zero, power is the ratio of the final K.E. and time, which is equal to 57870.37 Watt or 77.574 H.P.