# Suppose you calculate that the magnitude of the total force on hook 1 is 5.94 N. Now, you measure...

## Question:

Suppose you calculate that the magnitude of the total force on hook 1 is 5.94 N. Now, you measure the total mass of hook 1 to be 594 g. Find the percent difference between these values.

## Calculating the Percent Difference of Two Values:

Percent difference is the ratio of the difference between two values to their average and is expressed in percent form. This is one way of describing how two related values deviate from each other.

## Answer and Explanation: 1

The percent difference of the two mass values of the hook is 2 %.

If the calculated total force magnitude is {eq}F = 5.94\ \rm{N} {/eq}, then the mass of the hook is given by:

{eq}\displaystyle m = \frac{F}{g} \\ {/eq}

where {eq}g = 9.8\ \rm{\frac{m}{s^2}} {/eq} is the acceleration due to gravity.

Solving for the mass of the hook:

{eq}\displaystyle m = \frac{5.94\ \rm{N}}{9.8\ \rm{\frac{m}{s^2}} } \\ m = 0.606\ \rm{kg} \ \ \text{ or } {/eq}

{eq}m = 606\ \rm{g} {/eq}

When measured, the obtained value of mass is {eq}m_1 = 594\ \rm{g}. {/eq}

The percentage difference {eq}PD {/eq} of the two values of masses {eq}m \text{ and } m_1 {/eq} can be calculated using the equation:

{eq}\displaystyle PD = \frac{|m - m_1|}{\frac{m + m_1}{2}} \times 100 \% {/eq}

Substituting the masses {eq}m = 606\ \rm{g}\ \text{ and } m_1 = 594\ \rm{g}: {/eq}

{eq}\displaystyle PD = \frac{|606\ \rm{g} - 594\ \rm{g}|}{\frac{606\ \rm{g} + 594\ \rm{g}}{2}} \times 100 \% {/eq}

{eq}\displaystyle PD = 2.00 \% {/eq}

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Chapter 5 / Lesson 3In this lesson, we'll learn how simple interest can earn (or cost) you money. We'll also look at percent change problems to better understand how discounts and markups work.