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3 years ago

Is it possible to do work on an object without changing the kinetic energy of the object? Now Why?

Physics

1 answer:

Murrr4er [49]3 years ago

8 0

Answer:

(a) Yes, it is possible by raising the object to a greater height without acceleration.

Explanation:

The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in kinetic energy requires a change in velocity.

If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.

If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).

Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised to a greater height without acceleration.

Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".

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Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

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