Answer:
Factor by which kinetic energy increase = 4 times
Step-by-step explanation:
Given,
[tex]=\dfrac{35\times 1609.34}{3600}\ m/s[/tex]
= 15.64 m/s
Initial kinetic energy of the car is given by,
[tex]k_1\ =\ \dfrac{1}{2}.m.v_1^2[/tex]
[tex]=\ \dfrac{1}{2}\times 1500\times (15.64)^2\ joule[/tex]
= 183606.46 J
[tex]=\dfrac{70\times 1609.34}{3600}[/tex]
= 31.29 m/s
So, final kinetic energy of car is given by
[tex]k_2\ =\ \dfrac{1}{2}.m.v_2^2[/tex]
[tex]=\ \dfrac{1}{2}\times 1500\times (31.29)^2[/tex]
= 734425.84 J
Now, the ratio of final to initial kinetic energy can be given by,
[tex]\dfrac{k_2}{k_1}=\ \dfrac{734425.84}{183606.46}[/tex]
[tex]=>\ k_2\ =\ 4k_1[/tex]
Hence, the kinetic energy will increase by 4 times.