# A model train, with a mass of #9 kg#, is moving on a circular track with a radius of #15 m#. If the train's kinetic energy changes from #96 j# to #48 j#, by how much will the centripetal force applied by the tracks change by?

The change in centripetal force is

The centripetal force is

The kinetic energy is

Therefore,

The variation of kinetic energy is

So,

The change in centripetal force is

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To find the change in centripetal force applied by the tracks, you can use the formula for centripetal force: ( F_c = \frac{{mv^2}}{r} ), where ( m ) is the mass of the object, ( v ) is the velocity, and ( r ) is the radius of the circular track. Since kinetic energy is given by ( KE = \frac{1}{2}mv^2 ), you can use the given kinetic energy values to find the corresponding velocities. Then, you can calculate the centripetal force using these velocities and the given radius. Finally, find the difference between the initial and final centripetal forces to determine the change.

First, solve for the initial velocity (( v_1 )) and final velocity (( v_2 )) using the given kinetic energy values:

Initial kinetic energy: ( KE_1 = 96 ) J ( KE_1 = \frac{1}{2}mv_1^2 ) ( v_1^2 = \frac{2 \times KE_1}{m} ) ( v_1^2 = \frac{2 \times 96}{9} ) ( v_1^2 = \frac{192}{9} ) ( v_1^2 \approx 21.33 ) ( v_1 \approx \sqrt{21.33} ) ( v_1 \approx 4.62 ) m/s

Final kinetic energy: ( KE_2 = 48 ) J ( KE_2 = \frac{1}{2}mv_2^2 ) ( v_2^2 = \frac{2 \times KE_2}{m} ) ( v_2^2 = \frac{2 \times 48}{9} ) ( v_2^2 = \frac{96}{9} ) ( v_2^2 \approx 10.67 ) ( v_2 \approx \sqrt{10.67} ) ( v_2 \approx 3.27 ) m/s

Now, calculate the initial and final centripetal forces:

Initial centripetal force (( F_{c1} )): ( F_{c1} = \frac{mv_1^2}{r} ) ( F_{c1} = \frac{9 \times (4.62)^2}{15} ) ( F_{c1} \approx \frac{9 \times 21.33}{15} ) ( F_{c1} \approx \frac{192.03}{15} ) ( F_{c1} \approx 12.80 ) N

Final centripetal force (( F_{c2} )): ( F_{c2} = \frac{mv_2^2}{r} ) ( F_{c2} = \frac{9 \times (3.27)^2}{15} ) ( F_{c2} \approx \frac{9 \times 10.67}{15} ) ( F_{c2} \approx \frac{96.03}{15} ) ( F_{c2} \approx 6.40 ) N

Finally, find the change in centripetal force: ( \Delta F_c = F_{c2} - F_{c1} ) ( \Delta F_c = 6.40 , \text{N} - 12.80 , \text{N} ) ( \Delta F_c = -6.40 , \text{N} )

So, the change in centripetal force applied by the tracks is -6.40 N.

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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.

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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