Question

As the parachute-spacecraft system decelerates it falls through a vertical distance of 49m and loses 2.2 x 10 to the power of 5 J of kinetic energy.

During this time , 3,3 x 10 to the power of 5 J of energy is transferred from the system to the atmosphere.The total mass of the system is 610 kg 

Calculate the acceleration due to gravity as it falls through this distance

Answer 1

Explanations & Calculations

• Let's take that the parachute exerts some constant retarding force on the falling system hence it experiences some constant deceleration $\small a$.
• Then apply $\small v^2 = u^2 +2as$ downward for the motion of the system.

$\qquad\qquad \begin{aligned} \small \downarrow v^2&=\small u^2+2a\times49\\ \small a&=\small \frac{v^2-u^2}{98} \end{aligned}$

• Then from the data about the kinetic energy loss during this period we get,

$\qquad\qquad \begin{aligned} \small 2.2\times10^5&=\small E_{initial}-E_{final}\\ &=\small \frac{1}{2}M(u^2-v^2)\\ \small u^2-v^2&=\small \frac{2}{M}(2.2\times10^5) \end{aligned}$

• Substituting this in the previous equation, we get the result

$\qquad\qquad \begin{aligned} \small a&=\small -\frac{2(2.2\times10^5)}{98M}\\ &=\small -\frac{2(2.2\times10^5)}{98\times610}\\ &=\small \bold{-7.36\,ms^{-2}} \end{aligned}$

• Total energy loss is the sum of the two energy losses stated in the question.
• We cannot say that the total energy loss is all the mechanical energy loss.
• If that is, then the net force could be found using energy conservation & then using F =ma, the deceleration could be calculated
• Therefore, that $\small 3.3\times10^5J$ of energy loss does not seem to be useful in the calculations involved in this question.