The force applied against an object moving horizontally on a linear path is described by #F(x)=x^2-3x + 3 #. By how much does the object's kinetic energy change as the object moves from # x in [ 0 , 1 ]#?

Answer 1

Newton's second law of motion:

#F=m*a#

Definitions of acceleration and velocity:

#a=(du)/dt#

#u=(dx)/dt#

Kinetic energy:

#K=m*u^2/2#

Answer is:

#ΔK=11/6# #kg*m^2/s^2#

The second law of motion by Newton:

#F=m*a#
#x^2-3x+3=m*a#
Substituting #a=(du)/dt# does not help with the equation, since #F# isn't given as a function of #t# but as a function of #x# However:
#a=(du)/dt=(du)/dt*(dx)/dx=(dx)/dt*(du)/dx#
But #(dx)/dt=u# so:
#a=(dx)/dt*(du)/dx=u*(du)/dx#

We get a differential equation by substituting into the existing equation:

#x^2-3x+3=m*u(du)/dx#
#(x^2-3x+3)dx=m*udu#
#int_(x_1)^(x_2)(x^2-3x+3)dx=int_(u_1)^(u_2)m*udu#
The two speeds are unknown but the positions #x# are known. Also, mass is constant:
#int_(0)^(1)(x^2-3x+3)dx=m*int_(u_1)^(u_2)udu#
#[x^3/3-3x^2/2+3x]_0^1=m*[u^2/2]_(u_1)^(u_2)#
#(1^3/3-3*1^2/2+3*1)-(0^3/3-3*0^2/2+3*0)=m*(u_2^2/2-u_1^2/2)#
#11/6=m*u_2^2/2-m*u_2^2/2#
But #K=m*u^2/2#
#11/6=K_2-K_1#
#ΔK=11/6# #kg*m^2/s^2#
Note : the units are #kg*m^2/s^2# only if the distances given #(x in[0,1])# are in metres.
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Answer 2

To find the change in kinetic energy of the object as it moves from (x = 0) to (x = 1), we need to calculate the work done by the force (F(x)) over this interval, which will be equal to the change in kinetic energy.

The work done by a force is given by the formula:

[ W = \int_{x_1}^{x_2} F(x) , dx ]

Given that the force (F(x) = x^2 - 3x + 3) and the object moves from (x = 0) to (x = 1), we integrate (F(x)) with respect to (x) over the interval ([0, 1]):

[ W = \int_{0}^{1} (x^2 - 3x + 3) , dx ]

[ W = \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 3x \right]_{0}^{1} ]

[ W = \left( \frac{1^3}{3} - \frac{3(1)^2}{2} + 3(1) \right) - \left( \frac{0^3}{3} - \frac{3(0)^2}{2} + 3(0) \right) ]

[ W = \left( \frac{1}{3} - \frac{3}{2} + 3 \right) - (0 - 0 + 0) ]

[ W = \left( \frac{1}{3} - \frac{3}{2} + 3 \right) ]

[ W = \left( \frac{1}{3} - \frac{9}{6} + \frac{18}{6} \right) ]

[ W = \left( \frac{1}{3} + \frac{9}{6} \right) ]

[ W = \left( \frac{2}{6} + \frac{9}{6} \right) ]

[ W = \frac{11}{6} ]

The change in kinetic energy ((\Delta KE)) is equal to the work done:

[ \Delta KE = W = \frac{11}{6} ]

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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.

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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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