Question

A 10-kg block on a rough horizontal surface is attached to a light spring (force constant = 1400 N/m). The block is pulled 10.0 cm to the right from its equilibrium position and released from rest. The frictional force between the block and surface has a magnitude of 30 N. What is the kinetic energy of the block as it passes through its equilibrium position?

Answer 1

To develop this problem it is necessary to apply the concepts related to the conservation of Energy. In this case the definition concerning kinetic energy from the simple harmonic movement.

From the conservation of energy we know that the kinetic energy would be conserved through the work done by the frictional force and the simple harmonic potential energy, in other words:

$KE = PE_s +W_f$

$KE = \frac{1}{2}kA^2 + W_f$

Where,

$KE =$ Kinetic Energy

$PE_s =$Potential Harmonic Simple Energy

$W_f =$ Work made by friction.

Our values are given as,

$m = 10Kg  \rightarrow$ mass

$k = 1400N/m \rightarrow$ Spring constant

$A = 0.1m \rightarrow$Amplitude

$f_f = 30N \rightarrow$ Frictional Force

Replacing we have,

$KE = \frac{1}{2}kA^2 + W_f$

$KE =\frac{1}{2} 1400 * 0.1^2 + ( - 30 * 0.1)$

$KE = 4 J$

Therefore the Kinetic Energy of the block as it passes through its equlibrium position is 4J.