Question

A force of 300 N is used to stretch a horizontal spring with a 0.5 kg block attached to it by 0.25 m. The block is released from rest and undergoes simple harmonic motion. Calculate 2.4.1 Spring constant (2) 2.4.2 Mechanical energy applied (2) 2.4.3 Velocity when x = 15 cm (3)

Answer 1

Explanations & Calculations

• As long as the spring is kept stretched: in this case by 0.25m with the 300N force, the force generated within the spring as a counterforce to the applied force is the same.
• Therefore, the equation for a spring, $\small F=kx$ is applicable.
• Then,

$\qquad\qquad \begin{aligned} \small F&=\small 300N=k\times 0.25m\\ \small k&=\small \frac{300N}{0.25m}=\bold{1200\,Nm^{-1}} \end{aligned}$

• Energy stored initially in the spring is the applied energy that can be calculated by the equation $\small E=\frac{1}{2}kx^2$.
• Then,

$\qquad\qquad \begin{aligned} \small E&=\small \frac{1}{2}\times1200\,Nm^{-1}\times(0.25m)^2\\ &=\small \bold{37.5\,J} \end{aligned}$

• By the energy conservation between the starting point and the given point (measured 0.15m from the unstretched position: relaxed spring), the speed at that point can be calculated.

$\qquad\qquad \begin{aligned} \small 37.5J&=\small \frac{1}{2}kx_1^2+\frac{1}{2} mv^2\\ &=\small \frac{1}{2}\times1200\times(0.15)^2+\frac{1}{2}\times0.5kg\times v^2\\ \small v^2&=\small 96m^2s^{-2}\\ &=\small \bold{9.798ms^{-1}} \end{aligned}$