A model train, with a mass of #6 kg#, is moving on a circular track with a radius of #1 m#. If the train's kinetic energy changes from #27 j# to #36 j#, by how much will the centripetal force applied by the tracks change by?

Question

A model train, with a mass of #6 kg#, is moving on a circular track with a radius of #1 m#.  If the train's kinetic energy changes from #27 j# to #36 j#, by how much will the centripetal force applied by the tracks change by?

Charlotte Aaron · Accepted Answer

18 N

Let'assume Kinetic energy in the 1st case is K1 and in 2nd case K2 So,#(K2)/(K1)# = # (m/m) ((v2)^2 / (v1)^2)# so, ratio of square of v2 and v1 will be #36/27# Now, Centripetal force F is given as #m(v^2)/r# As,mass and radius are unchanged so we can say, #(F2)/(F1)# #prop# #((v2)^2 / (v1)^2)# so #(F2)/(F1)# = #36/27# hence change in force #4/3# rd of F1 Initial Centripetal force was #(6*88)/1# N i.e 54 N ( velocity calculated using 0.5 m #v^2# = 27) Now it has become 4/3 rd i.e 72 so change occurred by (72-54)=18 N

Isabella Alvarez · Answer

The change in centripetal force is #=18N#

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 mass is #m=6kg#

The radius of the track is #r=1m#

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#

#=2/1*(36-27)N#

#=18N#

Julia Aguilar · Answer

undefined To find the change in centripetal force, we can use the equation for centripetal force:

$F_c = \frac{mv^2}{r}$

Given that the mass of the train ($m$) is 6 kg, the radius ($r$) of the circular track is 1 m, and the initial kinetic energy is 27 J, we can find the initial velocity ($v_i$) using the formula for kinetic energy:

$KE = \frac{1}{2}mv^2$

Solving for $v_i$, we have:

$27 = \frac{1}{2} \times 6 \times (v_i)^2$
$v_i = \sqrt{\frac{2 \times 27}{6}}$
$v_i = \sqrt{9}$
$v_i = 3 \, \text{m/s}$

Now that we have the initial velocity, we can find the initial centripetal force ($F_{c_i}$):

$F_{c_i} = \frac{6 \times (3)^2}{1}$
$F_{c_i} = 54 \, \text{N}$

Next, we find the final velocity ($v_f$) using the final kinetic energy:

$36 = \frac{1}{2} \times 6 \times (v_f)^2$
$v_f = \sqrt{\frac{2 \times 36}{6}}$
$v_f = \sqrt{12}$
$v_f = 2\sqrt{3} \, \text{m/s}$

Now that we have the final velocity, we can find the final centripetal force ($F_{c_f}$):

$F_{c_f} = \frac{6 \times (2\sqrt{3})^2}{1}$
$F_{c_f} = \frac{6 \times 12}{1}$
$F_{c_f} = 72 \, \text{N}$

Finally, we find the change in centripetal force ($\Delta F_c$):

$\Delta F_c = F_{c_f} - F_{c_i}$
$\Delta F_c = 72 - 54$
$\Delta F_c = 18 \, \text{N}$

Therefore, the change in centripetal force applied by the tracks is 18 N.