The force applied against an object moving horizontally on a linear path is described by #F(x)= cospix+x #. By how much does the object's kinetic energy change as the object moves from # x in [ 0, 2 ]#?

Answer 1

I got: #DeltaK=2J#

We can use the Work-K.E. theorem that tells us that:

#W=DeltaK#

or that work done on the system is equal to the change in kinetic energy.

Here we need to evaluate the work of a variable force so we use the integral version of work as:

#W=int_(x_1)^(x_2)f(x)dx=int_0^2[cos(pix)+x]dx=int_0^2[cos(pix)]dx+int_0^2[x]dx=sin(pix)/pi+x^2/s|_0^2=sin(2pi)/pi+2^2/2-0=2J#

So:

#W=DeltaK=2J#
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Answer 2

To find the change in kinetic energy, integrate the force function F(x) over the interval [0, 2] to obtain the work done on the object. Then, use the work-energy principle, which states that the work done on an object equals the change in its kinetic energy. Finally, calculate the change in kinetic energy using the work done and the initial and final kinetic energies of the object.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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