## Alberta Free Tutoring And Homework Help For Physics 20

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### an object travelling in a straight line at a constant velocity with a kinetic energy of 2.0J. If the mass of the object was doubled and it's speed was tripled, determine the objects new kinetic energy

8 months ago

Kinetic Energy of an object is written as  $E=\frac{1}{2}mv^2$E=12 mv2

Given Kinetic energy for an object of mass (m) and velocity (v) =2J

$E=\frac{1}{2}mv^2=2J$E=12 mv2=2J

$=>mv^2=2\times2J=4J$=>mv2=2×2J=4J

Now, mass is doubled and velocity is tripled

Let’s define new mass as m1= 2m and new velocity as v1=3v

So, New Kinetic Energy(E1)

E1$\frac{1}{2}\left(m1\right)\left(v1\right)^2=\frac{1}{2}\left(2\times m\right)\left(3\times v\right)^2=9\times mv^2$12 (m1)(v1)2=12 (2×m)(3×v)2=9×mv2

As calculated above,  $mv^2=4J$mv2=4J

Therefore E1$9\times4J=36J$9×4J=36J

8 months ago

$m_2=2.m_1$m2=2.m1

$v_2=3.v_1$v2=3.v1

$E_{k1}=\frac{1}{2}.m_1.v_1^2$Ek1=12 .m1.v12     =>    $.m_1.v_1^2=2.E_{k1}=2\times2.0=4.0$.m1.v12=2.Ek1=2×2.0=4.0 J

$?E_{k2}=\frac{1}{2}.m_2.v_2^2=\frac{1}{2}.\left(2m_1\right).\left(3v_1\right)^2=m_1.\left(9v_1^2\right)=9.m_1.v_1^2$?Ek2=12 .m2.v22=12 .(2m1).(3v1)2=m1.(9v12)=9.m1.v12

$E_{k2}=9\times4.0=36.0$Ek2=9×4.0=36.0 J

7 months ago

$E1=\frac{1}{2}\left(m1\right)\left(v1\right)^2=2J$E1=12 (m1)(v1)2=2J
$E2=\frac{1}{2}\left(m2\right)\left(v2\right)^2$E2=12 (m2)(v2)2
$m2=2\left(m1\right),v2=3\left(v1\right)$m2=2(m1),v2=3(v1)
$\frac{E2}{E1}=\frac{\frac{1}{2}\left(m2\right)\left(v2\right)^2}{\frac{1}{2}\left(m1\right)\left(v1\right)^2}=\frac{E2}{2}J$E2E1 =12 (m2)(v2)212 (m1)(v1)2 =E22 J
$E2=\frac{2\left(\frac{1}{2}\right)2\left(m1\right)\left(3\left(v1\right)\right)^2}{\frac{1}{2}\left(m1\right)\left(v1\right)^2}=2\left(2\right)\left(3\right)^2=36J$E2=2(12 )2(m1)(3(v1))212 (m1)(v1)2 =2(2)(3)2=36J