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Law of Conservation of Energy

Example:

A block with a mass of 2Kg is sliding on a horizontal frictionless surface with a velocity of 2.5 m/s. The block strikes a spring (K=4×10^3 N/m), Assume the impact is elastic (Both kinetic energy and momentum and conserved)

A.) What is the initial kinetic energy of the system?

B.) What is the kinetic energy of the system when the spring is compressed to x=1cm?

C.) How fast is the block moving at this point?

Known

$m_b=2Kg$

$V_1=2.5 m/5$

$K=4x^3 N/m$

$x=1cm = 0.01m$

Find

$KE_1= -? 5$

$KE_2=? 5$

$V_2=?$

Draw a Picture

Draw two energy states

Find energy at these states

$E_1 =KE1+SE1 E_2=KE2+SE2$

$E_1=\frac{1}{2}\cdot m_b (V_1)^2 = \frac{1}{2} \cdot 2kg(2.5 \frac{m}{s})^2 = 6.255$

Set the equations equals

energy at state 1 = energy at state 2

KE1=KE2+SE2

Solve for KE2

KE2=KE1-SE2

(6.255)- $\frac{1}{2}K\cdot x^2$

6.255-( $\frac{1}{2}\cdot 4 \cdot 10^3 \frac{N}{m} \cdot(0.01)^2$ )=KE2=6.05J

Solve for $V_2$ from KE2

KE2= $\frac{1}{2} \cdot m_b \cdot (V_2)^2$

Solve for $V_2$

$(\frac{2\cdotKE_2}{m_b})^{1/2} = 2.45\frac{m}{s}=V_2$