A rocket of mass 2 times 10^5 kg starts from rest at ground level and accelerates vertically...

Question:

A rocket of mass {eq}\rm 2 \times 10^5\ kg {/eq} starts from rest at ground level and accelerates vertically upward at {eq}\rm 4\ m/s^2 {/eq}. What is the instantaneous power of the engine when the rocket is at a height of 312.5 m above the ground level?

a. {eq}\rm 25 \times 10^8\ W {/eq}
b. {eq}\rm 0.4 \times 10^8\ W {/eq}
c. {eq}\rm 1.38 \times 10^8\ W {/eq}
d. {eq}\rm 50 \times 10^8\ W {/eq}
e. {eq}\rm 98 \times 10^8\ W {/eq}

Kinematic Equations:

We use kinematic equations when we are finding certain parameters for the movement of an object. As we may know, there are various kinematic equations that we can utilize, and we would know which one is appropriate when we consider the desired and the given parameters for a problem.

Answer and Explanation: 1

We determine the power, P, applied by the rocket at the given instant with the known conditions. We take note that the power can be expressed as:

{eq}\displaystyle P = Fv {/eq}, where:

  • F is the applied force
  • v is the instantaneous velocity


We can acquire F by considering the acceleration, a, and the mass, m, of the rocket such that:

{eq}\displaystyle F = ma {/eq}


On the other hand, we acquire v by applying the following kinematic equation:

{eq}\displaystyle v^2= v_0^2 + 2ad {/eq}


Taking the positive root, we have:

{eq}\displaystyle v = \sqrt{v_0^2 + 2ad } {/eq}, where:

  • {eq}\displaystyle v_0 {/eq} is the initial velocity
  • d is the current distance that has been traveled


We'll use the following values for the variables:

  • {eq}\displaystyle m = 2\times 10^5\ kg {/eq}
  • {eq}\displaystyle a = 4\ m/s^2 {/eq}
  • {eq}\displaystyle v_0= 0\ m/s {/eq}
  • {eq}\displaystyle d = 312.5\ m {/eq}


We proceed with the solution:

{eq}\begin{align} \displaystyle P &= Fv \\[0.3cm] &= ma\cdot \sqrt{v_0^2 + 2ad } &= (2\times 10^5\ kg)(4\ m/s^2)\cdot \sqrt{(0\ m/s)^2 + 2(4\ m/s^2)(312.5\ m)} \\[0.3cm] &= \boxed{4\times 10^7\ W} \end{align} {/eq}


Therefore, the answer is found to be b. {eq}\bf 0.4\times 10^{8}\ W {/eq}.


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