the speed of a particle doubles and then doubles again because a net external force acts on it. Does the net f

Question

the speed of a particle doubles and then doubles again because a net external force acts on it. Does the net force do more work during the first or the second doubling? explain

Answer 1

Work is the integral of force over a differential distance. If teh force is independent of position then you get the simple formula:

W = Fx where F =force, x = distance, and W = work

Now W = change in kinetic energy so in the first doubleing of speed, the work done is

W = 1/2 m*(v1^2 - v0^2) = 1/2 m*(4v0^2 - v0^2) = 3/2 m v0^2

where v0 = speed at start of problem

Now second doubling gives work of

W2 = 1/2 m*(v2^2 - v1^2) = 1/2 m (4v1^2 - v1^2) = 3/2 m v1^2

But v1 = 2v0 so W2 = 3/2 m 4v0^2 = 6 mv0^2

So force does more work for the second doubling

Answer 2

A chart with each one giving rise to the next in line.
force->acceleration->velocity
Theoretically, the force could remain constant.
However, in a practical sense, more force usually
is required to overcome resistive forces as
velocities increase. For example, a free falling
object can attain a terminal velocity in which
the gravitational force is not sufficient to overcome
the resistive force of air resistance. Yes, more
force would be required to farther increase the
falling objects speed(which would no longer
be free falling). However, you specifically state
the word "work", and it is possible that the
work done does not need to increase. Perhaps,
the prior input work is recycled as the system
continues to receive new input, thereby attaining
an amount of power which increases with time
while input work remains constant.

Answer 3

Let V be the initial velocity and next velocity is 2V and again it doubles to 4V

Work done by the force during fist phase is

W1 = [m/2] [4V^2 - V^2] = [m/2] [3V^2]

Work done by the force during second phase is

W2 = [m/2] [16V^2 - 4V^2] = [m/2] [12V^2]

W2/ W1 = 4.

The force is doing 4 times the work than it done during the first phase
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Answer 4

second doubling