Robert has just bought a new model rocket, and is trying to measure its flight characteristics....

Question:

Robert has just bought a new model rocket and is trying to measure its flight characteristics. The rocket engine package claims that it will maintain a constant thrust of 14.1 N until the engine is used up. Robert launches the rocket on a windless day, so that it travels straight up, and uses his laser range-finder to measure that the height of the rocket when the engine cuts off is 16.8 m. He also measures the rocket's peak height, which is 21.3 m. If the rocket has a mass of 0.763 kg, how much work is done by the drag force on the rocket during its ascent?

Law of Conservation of Energy:

One of the most fundamental laws of physics is the law of conservation of energy. It states that energy can only be transformed from one form to another, or distributed in various forms, but can not be created or destroyed.

The total energy of a system and its surrounding is constant, irrespective of any interaction happening between the system and its surrounding.

If one form of the energy of a system is decreasing, then this energy must simultaneously appear in the system in some other form.

• The mass of the rocket is: {eq}m=0.7763\;\rm kg {/eq}
• The upward force provided by the rocket up to height {eq}h_1=16.8\;\rm m {/eq}, is {eq}F=14.1\;\rm N {/eq}
• The maximum height reached by rocket is: {eq}h_2=21.3\;\rm m {/eq}

The work done by a force {eq}F {/eq} in displacing an object by {eq}x {/eq} units is given by:

{eq}W=Fx {/eq}

The work done by the upward force (thrust) on the rocket is:

{eq}W=14.1\times 16.8=236.88\;\rm J {/eq}

According to the Work-Energy principle, the work {eq}W {/eq} done on an object is equal to the change in its kinetic energy {eq}\Delta K {/eq}

{eq}W=\Delta K {/eq}

Therefore, the kinetic energy of the rocket at {eq}h_1 {/eq} is:

{eq}K_1=W=236.88\;\rm J {/eq}

At this height, the engine cuts off. After this height, the rocket continues to move upward until all the kinetic energy of the rocket is used up in increasing the gravitational potential energy, {eq}U {/eq}, of the rocket and doing work, {eq}W {/eq}, against the drag forces. That is:

{eq}K_1=\Delta U+W\;\;\;\;\;\;\;\text{(Equation 1)} {/eq}

The change in the gravitational potential energy of an object of mass, {eq}m {/eq} when its height is changed by {eq}\Delta x {/eq}, is:

{eq}\Delta U=mg\Delta h {/eq}

where {eq}g=9.81\;\rm m/s^2 {/eq} is the acceleration due to gravity.

The change in the height of the rocket after the engine cuts off, is:

{eq}\Delta h=21.3-16.8=4.5\;\rm m {/eq}

Therefore,

{eq}\Delta U=0.763\times 9.81\times 4.5=33.68\;\rm J {/eq}

On plugging the value of {eq}\Delta U {/eq}, and {eq}K_1 {/eq} in eq (1), we have:

{eq}236.88=33.68+W\\ \Rightarrow \boxed{W=203.2\;\rm J} {/eq} 