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 #18 j#, by how much will the centripetal force applied by the tracks change by?

Question

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 #18 j#, by how much will the centripetal force applied by the tracks change by?

Benjamin Asiedu · Accepted Answer

The centripetal force changes by #=4N#

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*(36-18)N#

#=4N#

Xavier Adame · Answer

undefined To find the change in centripetal force, we can use the relationship between kinetic energy and centripetal force.

The kinetic energy of an object moving in a circular path is given by:

$$ KE = \frac{1}{2} m v^2 $$

The centripetal force acting on an object moving in a circular path is given by:

$$ F_c = \frac{m v^2}{r} $$

Given:
Initial kinetic energy ($ KE_1 $) = 36 J
Final kinetic energy ($ KE_2 $) = 18 J
Mass of the train ($ m $) = 12 kg
Radius of the circular track ($ r $) = 9 m

Using the equation for kinetic energy:

$$ KE_1 = \frac{1}{2} m v_1^2 $$

$$ KE_2 = \frac{1}{2} m v_2^2 $$

Solving for initial and final velocities:

$$ v_1 = \sqrt{\frac{2 \times KE_1}{m}} $$

$$ v_2 = \sqrt{\frac{2 \times KE_2}{m}} $$

Now, calculate initial and final velocities:

$$ v_1 = \sqrt{\frac{2 \times 36}{12}} = \sqrt{6} \approx 2.45 \, m/s $$

$$ v_2 = \sqrt{\frac{2 \times 18}{12}} = \sqrt{3} \approx 1.73 \, m/s $$

Now, calculate the initial and final centripetal forces:

$$ F_{c1} = \frac{m v_1^2}{r} $$

$$ F_{c2} = \frac{m v_2^2}{r} $$

Substitute the values:

$$ F_{c1} = \frac{12 \times (2.45)^2}{9} \approx 8.08 \, N $$

$$ F_{c2} = \frac{12 \times (1.73)^2}{9} \approx 4.16 \, N $$

The change in centripetal force is:

$$ \Delta F_c = F_{c2} - F_{c1} $$

$$ \Delta F_c = 4.16 \, N - 8.08 \, N $$

$$ \Delta F_c \approx -3.92 \, N $$

Therefore, the centripetal force applied by the tracks decreases by approximately 3.92 N.