# A 5.0 kg mass hanging from a spring scale is slowly lowered onto a vertical spring. a) What...

## Question:

A 5.0 kg mass hanging from a spring scale is slowly lowered onto a vertical spring.

a) What does the spring scale read just before the mass touches the lower spring? (N)

b) The scale reads 18 N when the lower spring has been compressed by 2.0 cm. What is the value of the spring constant for the lower spring? (N/m)

c) At what compression length will the scale read zero? (cm)

## Hooke's Law:

The restoring force of a spring is given by Hooke's law

{eq}F=-kx {/eq}

Where,

- {eq}k {/eq} is the spring constant;

- {eq}x {/eq} is the compression distance.

Take note that the restoring force is negative and always points towards the equilibrium.

## Answer and Explanation: 1

Given the following quantities:

- mass {eq}m=5.0\text{ kg} {/eq};

- acceleration due to gravity {eq}9.8\text{ m/s}^2 {/eq}

Let the +y-axis be the positive direction. Therefore, downward compression is negative and the restoring force is upward and positive.

a) Just before the mass touches the lower spring, the scale reading is equivalent to the magnitude of the weight of the mass.

{eq}\begin{align} w&=mg\\ &=\left(5.0\text{ kg}\right)\left(9.8\text{ m/s}^2\right)\\ w&=49\text{ N} \end{align} {/eq}

b) When the lower spring is compressed downwarf by {eq}y=-2.0\text{ cm}=-0.020\text{ m} {/eq} the apparent weight of the mass on the scale reads {eq}w'=18\text{ N} {/eq}.

Notice that the apparent weight is equivalent to the difference between the actual weight and the restoring force {eq}F {/eq} of the lower spring.

{eq}w'=w-F {/eq}

and

{eq}\begin{align} F&=w-w'\\ &=49\text{ N}-18\text{ N}\\ F&=31\text{ N} \end{align} {/eq}

From Hooke's law,

{eq}F=-ky {/eq}

Solve for the spring constant {eq}k {/eq},

{eq}\begin{align} k&=-\frac{F}{y}\\ &=-\frac{31\text{ N}}{-0.020\text{ m}}\\ k&=1550\text{ N/m} \end{align} {/eq}

c) The reading in the scale becomes zero when the magnitude of the restoring force is equal to the weight of the mass

{eq}F=49\text{ N} {/eq}

Solving for the compression using Hooke's law,

{eq}F=-ky {/eq}

and

{eq}\begin{align} y&=-\frac{F}{k}\\ &=-\frac{49\text{ N}}{1550\text{ N/m}}\\ y&\approx -0.031\text{ m} \end{align} {/eq}

or

{eq}y\approx -3.1\text{ cm} {/eq}

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Chapter 4 / Lesson 19After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.