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Chapter 7: Problem 917

When a \(4 \mathrm{~kg}\) mass is hung vertically on a light spring that obeys Hook's law, the spring stretches by \(2 \mathrm{~cm}\) what will be the work required to be done by an external agent in stretching this spring by $5 \mathrm{~cm}$ ? (A) \(4.900\) Joule (B) \(2.450\) Joule (C) \(0.495\) Joule (D) \(0.245\) Joule

Short Answer

Expert verified
The work required to be done by an external agent in stretching the spring by 5 cm is approximately 2.0621 Joules, and the closest option is (B) 2.450 Joule.

Step by step solution

01

Identify the Given Information and Knowns

According to the problem, we know the following information: Mass, m = 4 kg Initial spring stretch, x1 = 2 cm (0.02 m, converted from cm to meters) Additional stretch, x2 = 5 cm (0.05 m, converted from cm to meters) From Hooke's law, F = kx, we need to find the spring constant, 'k', and then calculate the work done, W.
02

Calculate the Spring Constant

From Hooke's law, F = kx, and we know the force acting on the spring is the gravitational force due to the mass (m) hung on the spring, which is F = mg. Now, substitute this force and the initial stretch into Hooke's law: mg = kx1 4 kg × 9.81 m/s² = k × 0.02 m Now solve for k: k = (4 kg × 9.81 m/s²) / 0.02 m k = 1962 N/m
03

Calculate the Work Done

To find the work done, we use the formula for work done on a spring, which is W = (1/2)k(x2² - x1²). Plug in the values for k, x1, and x2: W = (1/2)(1962 N/m)(0.05 m² - 0.02 m²) W = (1/2)(1962 N/m)(0.0025 m² - 0.0004 m²) W = (1/2)(1962 N/m)(0.0021 m²) W = 2.0621 J The work required to be done by an external agent in stretching the spring by 5 cm is approximately 2.0621 Joules. The closest answer among the given options is (B) 2.450 Joule.

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