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Chapter 4: Problem 421

A \(60 \mathrm{~kg}\) JATAN with \(10 \mathrm{~kg}\) load on his head climbs 25 steps of \(0.20 \mathrm{~m}\) height each. What is the work done in climbing ? \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(5 \mathrm{~J}\) (B) \(350 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(3500 \mathrm{~J}\)

Short Answer

Expert verified
The work done in climbing is 3500 J. The correct answer is (D) \(3500 \mathrm{~J}\).

Step by step solution

01

1. Identify the given variables.

We are given the following variables: - Mass of JATAN (m1) = 60 kg - Mass of load (m2) = 10 kg - Number of steps (n) = 25 - Height of each step (h) = 0.20 m - Gravitational acceleration (g) = 10 m/s²
02

2. Calculate the total mass being lifted.

The total mass being carried by the person as he climbs the stairs is the sum of the JATAN's own mass and the mass of the load. Therefore, we have: Total mass (m) = m1 + m2 m = 60 kg + 10 kg = 70 kg
03

3. Determine the total height climbed by the person.

To find the total height climbed by the person, multiply the number of steps by the height of each step: Total height (H) = n * h H = 25 * 0.20 m = 5 m
04

4. Calculate the change in potential energy.

The change in potential energy (ΔU) can be calculated using the formula: ΔU = m * g * H ΔU = 70 kg * 10 m/s² * 5 m = 3500 J
05

5. Determine the work done in climbing.

According to the work-energy principle, the work done (W) is equal to the change in potential energy: W = ΔU W = 3500 J So the work done in climbing is 3500 J. The correct answer is (D) \(3500 \mathrm{~J}\).

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