Question

A horizontal spring attached to a wall has a force constant of k=850N/m. A block of mass $\mathit{m}\mathbf{=}\mathbf{1}\mathbf{.00}\mathbf{}\mathbf{}\mathit{k}\mathit{g}$is attached to the spring and rests on a frictionless, horizontal surface as in Figure P8.59. (a) The block is pulled to a position ${\mathit{x}}_{\mathbf{i}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$from equilibrium and released. Find the elastic potential energy stored in the spring when the block is 6.00 cm from equilibrium and when the block passes through equilibrium. (b) Find the speed of the block as it passes through the equilibrium point. (c) What is the speed of the block when it is at a position $\frac{{\mathbf{x}}_{\mathbf{i}}}{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.00}\mathbf{}\mathbf{}\mathit{m}$? (d) Why isn’t the answer to part (c) half the answer to part (b)?

Short answer

(a) The elastic potential energy stored in the spring is$U=1.53J$.

(b) The speed of the block as it passes through the equilibrium point is 1.75m/s.

(c) The speed of the block when it is at a position $\frac{x_i}{2}=3.00m$is 1.51 m/s

(d) As the relation between the velocity and the displacement is not linear, so by reducing the displacement to half, the velocity does not reduce to half.

Step-by-step solution

Introduction

If the restoring force acting on a body is directly proportional to the displacement of the body from its mean position, then the motion of the body is called a simple harmonic motion. The elastic energy stored in the stretched string is mathematically presented as:

$\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$

Here:

U = Potential energy

k = Spring constant

x = Displacement from the mean position

Given Data

Mass of the body: m=1.0 Kg

Force constant of the spring: k=850 N/m

Displacement from the mean position: x=6 cm

Calculation for part (a)

The energy stored in the spring is the elastic potential energy.

For, $k=850N/m$and $x=0.600m$, the elastic energy is calculated as

$\begin{array}{c}U=\frac{1}{2}k{x}^2\\ =\frac{1}{2}\left(850\text{N/m}\right){\left(0.0600\text{m}\right)}^2\\ =1.53\text{J.}\end{array}$

At the equilibrium position x=0, so the value of potential energy is:

$\begin{array}{c}U=\frac{1}{2}k{x}^2\\ =\frac{1}{2}k\times {\left(0\right)}^2\\ =0\text{J.}\end{array}$

Calculation for part (b)

As the body is released from rest, apply the energy conservation to the block-spring system:

$\begin{array}{c}\Delta K+\Delta U=0\\ \frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2+\left(U_f-U_i\right)=0\\ \frac{1}{2}m{v}_f^2=-\left(U_f-U_i\right)\\ =\left(U_i-U_f\right).\end{array}$

$x_f=0,U_f=0$therefore, we have:

$\begin{array}{c}\frac{1}{2}m{v}_f^2=U_i-U_f\\ v_f=\sqrt{\frac{2\left(U_i-U_f\right)}{m}}\\ =\sqrt{\frac{2\left(1.53\text{J}\right)}{1.00\text{kg}}}\\ =1.75\text{m/s.}\end{array}$

Calculation for part (c)

From part (b) above$x_f=\frac{x_i}{2}=0.0300m$

$\begin{array}{c}U=\frac{1}{2}k{x}^2=\frac{1}{2}\left(\text{850N/m}\right){\left(0.0300\text{m}\right)}^2\\ =0.383\text{J}\end{array}$

$\begin{array}{l}\frac{1}{2}m{v}_f^2=U_i-U_f\\ v_f=\sqrt{\frac{2\left(U_i-U_f\right)}{m}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\\ =\sqrt{\frac{2\left(1.53\text{J}-0.383\text{J}\right)}{1.00\text{kg}}}\\ =\sqrt{\frac{2\left(1.15\text{J}\right)}{1.00\text{kg}}}=1.51\text{m/s.}\end{array}$

Calculation for part (d)

Using equation (2), we have:

$\begin{array}{c}{v}_f=\sqrt{\frac{2\left(\frac{1}{2}k{x}_f^2-\frac{1}{2}k{x}_i^2\right)}{m}}\\ =\sqrt{\frac{k\left(x_f^2-x_i^2\right)}{m}}.\end{array}$

As the above relation between the velocity and the displacement is not linear, therefore, on reducing displacement to half of the initial value, the velocity is not reduced to half.