Respuesta :
Answer:
The kinetic energy of car with mass 1500 kg and with speed of 35 miles/hour is KE=183598 J and when the car increases its speed to 70 miles/hour the kinetic energy changes by a factor of 4.
Step-by-step explanation:
The first step is to convert the speed miles/hour to m/s.
[tex]35\frac{miles}{hour} *\frac{1609.34 \>m}{1 \>miles}*\frac{1 \>hour}{3600 \> s}=15.646 \frac{m}{s}[/tex]
Next, the formula for the kinetic energy is
[tex]KE=\frac{1}{2} mv^{2}[/tex]
So input the values given:
[tex]KE=\frac{1}{2} (1500)(15.646)^{2}\\KE=750 \cdot (15.646)^{2}\\KE=183597.987 = 183598 \frac{kg \cdot \>m^{2}}{s^{2}} \\KE=183598 \>J[/tex]
Notice that the speed of 70 miles/hour is the double of 35 miles/hour so we can say that [tex]v_{2}=2v_{1}[/tex] and use the formula for the kinetic energy
[tex]KE_{2} =\frac{1}{2} m(v_{2}) ^{2}\\if \: v_{2}= 2v_{1}, then \:\\KE_{2} =\frac{1}{2} m(2v_{1}) ^{2}\\KE_{2} =\frac{1}{2} m4(v_{1})^{2}\\KE_{2} =4(\frac{1}{2} m(v_{1})^{2})\\We \:know \:that \:KE_{1} =\frac{1}{2} m(v_{1})^{2} so\\KE_{2} =4(KE_{1})[/tex]
We can see that when the car increases its speed to 70 miles/hour the kinetic energy changes by a factor of 4.
