A model train, with a mass of #12 kg#, is moving on a circular track with a radius of #9 m#. If the train's kinetic energy changes from #36 j# to #27 j#, by how much will the centripetal force applied by the tracks change by?

Answer 1

The change in centripetal force is #=2N#

The centripetal force is

#F=(mv^2)/r#

The kinetic energy is

#KE=1/2mv^2#

The variation of kinetic energy is

#Delta KE=1/2mv^2-1/2m u^2#
#=1/2m(v^2-u^2)#
The radius is #=9m#

The variation of centripetal force is

#DeltaF=m/r(v^2-u^2)#
#DeltaF=2m/r1/2(v^2-u^2)#
#=(2)/r*1/2m(v^2-u^2)#
#=(2)/r*Delta KE#
#=2/9*(36-27)N#
#=2N#
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Answer 2

The change in centripetal force applied by the tracks will be 9 N.

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Answer 3

The centripetal force ( F_c ) acting on an object moving in a circular path can be calculated using the formula:

[ F_c = \frac{mv^2}{r} ]

where ( m ) is the mass of the object, ( v ) is its velocity, and ( r ) is the radius of the circular path.

First, we need to find the initial velocity of the model train using its initial kinetic energy ( KE_1 ) and then the final velocity using its final kinetic energy ( KE_2 ). The kinetic energy ( KE ) is given by the formula:

[ KE = \frac{1}{2}mv^2 ]

Given that ( KE_1 = 36 , \text{J} ) and ( KE_2 = 27 , \text{J} ), we can set up the equations:

[ 36 = \frac{1}{2} \times 12 \times v_1^2 ] [ 27 = \frac{1}{2} \times 12 \times v_2^2 ]

Solving for ( v_1 ) and ( v_2 ):

[ v_1^2 = \frac{36 \times 2}{12} = 6 ] [ v_1 = \sqrt{6} \approx 2.45 , \text{m/s} ]

[ v_2^2 = \frac{27 \times 2}{12} = 4.5 ] [ v_2 = \sqrt{4.5} \approx 2.12 , \text{m/s} ]

Now, we can calculate the initial centripetal force ( F_{c1} ) and the final centripetal force ( F_{c2} ) using the formula ( F_c = \frac{mv^2}{r} ):

[ F_{c1} = \frac{12 \times (2.45)^2}{9} \approx 7.32 , \text{N} ] [ F_{c2} = \frac{12 \times (2.12)^2}{9} \approx 5.02 , \text{N} ]

The change in centripetal force is given by ( \Delta F_c = F_{c2} - F_{c1} ):

[ \Delta F_c = 5.02 , \text{N} - 7.32 , \text{N} = -2.3 , \text{N} ]

Therefore, the centripetal force applied by the tracks will decrease by approximately 2.3 N.

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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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