# When a 3.70-\mathrm{kg} object is hung vertically on a certain light spring that obeys Hooke's...

## Question:

When a {eq}3.70-\mathrm{kg} {/eq} object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches {eq}2.10\ \mathrm{cm} {/eq}.

(a) If the {eq}3.70-\mathrm{kg} {/eq} object is removed, how far will the spring stretch if a {eq}1.50-\mathrm{kg} {/eq} block is hung on it?

(b) How much work must an external agent do to stretch the same spring {eq}4.00\ \mathrm{cm} {/eq} from its unstretched position?

## Hooke's Law:

Hooke's law state that the amount of defection in spring is proportional to the weight applied to it. Force in spring is proportional to the displacement, and Work is done by spring is proportional to the square of the displacement.

## Answer and Explanation:

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Given data

• Initial mass of an object is: {eq}M = 3.70\;{\rm{kg}} {/eq}
• Extension of spring is: {eq}x = 2.10\;{\rm{cm}} =...

See full answer below.

#### Learn more about this topic: Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
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After 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.