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Chapter 4: Problem 463

A spring is compressed by \(1 \mathrm{~cm}\) by a force of \(4 \mathrm{~N}\). Find the potential energy of the spring when it is compressed by \(10 \mathrm{~cm}\) (A) \(2 \mathrm{~J}\) (B) \(0.2 \mathrm{~J}\) (C) \(20 \mathrm{~J}\) (D) \(200 \mathrm{~J}\)

Short Answer

Expert verified
The potential energy of the spring when it is compressed by 10 cm is \(2 \mathrm{~J}\). The correct answer is (A) \(2 \mathrm{~J}\).

Step by step solution

01

Apply Hooke's Law to find the spring constant

According to Hooke's Law, the force needed to compress a spring is proportional to its displacement (compression). Mathematically, it is represented as: \(F = kx\) where \(F\) is the force needed to compress the spring, \(k\) is the spring constant, and \(x\) is the displacement (the amount the spring is compressed). We are given that when the spring is compressed by 1 cm (0.01 m), the force required is 4 N. We can plug these values into Hooke's Law to find the spring constant, \(k\). \(4 \mathrm{~N} = k(0.01 \mathrm{~m})\)
02

Calculate the spring constant

To find the spring constant \(k\), we can simply divide the force by the displacement: \(k = \dfrac{4 \mathrm{~N}}{0.01 \mathrm{~m}}\) \(k = 400 \mathrm{~N/m}\) Now we have the spring constant, which is 400 N/m.
03

Apply the Potential Energy formula to find the potential energy when the spring is compressed by 10 cm

The potential energy stored in a compressed spring is given by the formula: PE = 0.5 × k × x^2 where PE is the potential energy, k is the spring constant, and x is the displacement (the amount the spring is compressed). We are given that the spring is compressed by 10 cm (0.1 m), and we have already found the spring constant, so now we can plug these values into the formula to find the potential energy. PE = 0.5 × 400 N/m × (0.1 m)^2
04

Calculate the potential energy

Now, we just need to compute the potential energy: PE = 0.5 × 400 N/m × (0.1 m)^2 PE = 0.5 × 400 N/m × 0.01 m^2 PE = 200 N/m × 0.01 m^2 PE = 2 J The potential energy of the spring when it is compressed by 10 cm is 2 J. The correct answer is (A) \(2 \mathrm{~J}\).

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Most popular questions from this chapter

An electric motor develops \(5 \mathrm{KW}\) of power. How much time will it take to lift a water of mass \(100 \mathrm{~kg}\) to a height of $20 \mathrm{~m}\( ? \)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(4 \mathrm{sec}\) (B) \(5 \mathrm{sec}\) (C) \(8 \mathrm{sec}\) (D) \(10 \mathrm{sec}\)

A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one third of its is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL (B) \([(\mathrm{MgL}) /(3)]\) (C) \([(\mathrm{MgL}) /(9)]\) (D) \([(\mathrm{MgL}) /(18)]\)

\(1 \mathrm{~kg}\) apple gives \(25 \mathrm{KJ}\) energy to a monkey. How much height he can climb by using this energy if his efficiency is \(40 \%\). (mass of monkey \(=25 \mathrm{~kg}\) and \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(20 \mathrm{~m}\) (B) \(4 \mathrm{~m}\) (C) \(30 \mathrm{~m}\) (D) \(40 \mathrm{~m}\)

The potential energy of a projectile at its highest point is (1/2) th the value of its initial kinetic energy. Therefore its angle of projection is (A) \(30^{\circ}\) (B) \(45^{\circ}\) (C) \(60^{\circ}\) (D) \(75^{\circ}\)

A metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of $36 \mathrm{~km} / \mathrm{h}$ has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)
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