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Chapter 4: Problem 26

A man lifts a 2.0 -kg stone vertically with his hand at a constant upward acceleration of \(1.5 \mathrm{m} / \mathrm{s}^{2} .\) What is the magnitude of the total force of the man's hand on the stone?

Short Answer

Expert verified
Answer: The magnitude of the total force of the man's hand on the stone is 22.62 N.

Step by step solution

01

Identify the forces acting on the stone

There are two forces acting on the stone: gravitational force (weight) acting downwards and the applied force by the man acting upwards.
02

Calculate the gravitational force (weight)

The gravitational force (weight) acting on the stone can be calculated using the formula: \(W = mg\), where \(m\) is the mass of the stone and \(g\) is the gravitational acceleration (\(9.81 \, \mathrm{m}/\mathrm{s}^{2}\)). Plugging in the values, we have \(W = (2.0\, \mathrm{kg})(9.81 \, \mathrm{m}/\mathrm{s}^{2}) = 19.62 \, \mathrm{N}\)
03

Calculate the net force using Newton's second law of motion

According to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration (\(F_{net} = ma\)). Using the given values, we have \(F_{net} = (2.0\, \mathrm{kg})(1.5 \, \mathrm{m}/\mathrm{s}^{2}) = 3.0 \, \mathrm{N}\)
04

Determine the applied force by the man

To find the magnitude of the force exerted by the man's hand on the stone, we need to combine the gravitational force and the net force. Since the stone is moving upward and the gravitational force acts downward, the applied force must be greater than the gravitational force. Thus, \(F_{applied} = F_{net} + W = 3.0 \, \mathrm{N} + 19.62 \, \mathrm{N} = 22.62 \, \mathrm{N}\)
05

Report the final answer

The magnitude of the total force of the man's hand on the stone is \(22.62 \, \mathrm{N}\).

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