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Chapter 4: Problem 496

\(1 \mathrm{~kg}\) 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) \(40 \mathrm{~m}\)

Short Answer

Expert verified
The monkey can climb 40 meters with the energy provided by the 1 kg apple, so the correct answer is (D) \(40 \mathrm{~m}\).

Step by step solution

01

Convert the given energy to work done by the monkey

The monkey has an efficiency of 40%, which means only 40% of the energy from the apple is converted into work done by the monkey. Therefore, we need to calculate the actual work done by the monkey: Work done = Efficiency × Energy from apple \( Work \, done = 0.40 \times 25 \, KJ \)
02

Convert work done to potential energy

The work done by the monkey is equal to the change in potential energy (PE) as it climbs. We can write the potential energy equation as: PE = Mass × Gravitational acceleration × Height In our case, the mass of the monkey is 25 kg, and gravitational acceleration is 10 m/s². We are interested in the height climbed, so we can rewrite the equation as: Height = PE / (Mass × Gravitational acceleration) Now let's calculate the height climbed by the monkey.
03

Calculate the height climbed

Using the values calculated in Step 1 and Step 2, we can now find the height climbed by the monkey: Height = \( \frac{Work \, done}{(Mass \, × \, Gravitational \, acceleration)} \) Height = \( \frac{0.40 \times 25 \, KJ}{(25 \, kg \, × \, 10 \, m/s^{2})} \) First, we need to convert KJ to J (1 KJ = 1000 J): Height = \( \frac{0.40 \times 25000 \, J}{(25 \, kg \, × \, 10 \, m/s^{2})} \) Now, we can calculate the height: Height = \( \frac{10000 \, J}{(250 \, kg \cdot m/s^{2})} \) Height = 40 m #Conclusion# The monkey can climb 40 meters with the energy provided by the 1 kg apple, so the correct answer is (D) \(40 \mathrm{~m}\).

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Most popular questions from this chapter

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of \(18 \mathrm{~cm}\) (Take \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(0.4 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(1.8 \mathrm{~m} / \mathrm{s}\) (D) \(0.6 \mathrm{~m} / \mathrm{s}\)

A ball dropped from a height of \(4 \mathrm{~m}\) rebounds to a height of $2.4 \mathrm{~m}$ after hitting the ground. Then the percentage of energy lost is (A) 40 (B) 50 (C) 30 (D) 600

If Wa, Wb, and Wc represent the work done in moving a particle from \(\mathrm{X}\) to \(\mathrm{Y}\) along three different path $\mathrm{a}, \mathrm{b}\(, and \)\mathrm{c}$ respectively (as shown) in the gravitational field of a point mass \(\mathrm{m}\), find the correct relation between \(\mathrm{Wa}, \mathrm{Wb}\) and \(\mathrm{Wc}\) (A) \(\mathrm{Wb}>\mathrm{Wa}>\mathrm{Wc}\) (B) \(\mathrm{Wa}<\mathrm{Wb}<\mathrm{Wc}\) (C) \(\mathrm{Wa}>\mathrm{Wb}>\mathrm{Wc}\) (D) \(\mathrm{Wa}=\mathrm{Wb}=\mathrm{Wc}\)

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) \(75^{\circ}\)

The coefficient of restitution e for a perfectly elastic collision is (A) 1 (B) 0 (C) \(\infty\) (D) \(-1\)
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