Law of Conservation of Energy


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?




V_1=2.5 m/5

K=4x^3 N/m

x=1cm = 0.01m


KE_1= -? 5

KE_2=? 5


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


Solve for KE2


(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