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Chapter 4: Problem 25

A man lifts a 2.0 -kg stone vertically with his hand at a constant upward velocity of \(1.5 \mathrm{m} / \mathrm{s} .\) 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 exerted by the man's hand on the stone is 19.62 N.

Step by step solution

01

Calculate gravitational force

To calculate the gravitational force acting on the stone, we can use the formula: \(F_g = m \times g\) where \(F_g\) is the gravitational force, \(m\) is the mass of the stone, and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{m} / \mathrm{s}^2\)). Given the mass of the stone as 2.0 kg, we can plug in the values and find the gravitational force. \(F_g = 2.0 \, \text{kg} \times 9.81 \, \frac{\text{m}}{\text{s}^2} = 19.62 \, \text{N}\)
02

Determine net force

Since the stone is moving upward with a constant velocity, the net force acting on it must be zero (according to Newton's first law). So, we have: \(\Sigma F_y = 0\)
03

Calculate the magnitude of the total force exerted by the man's hand

To find the magnitude of the total force exerted by the man's hand (\(F_h\)) on the stone, we can use the net force equation we wrote in Step 2: \(\Sigma F_y = F_h - F_g = 0\) Now we can solve for the force exerted by the man's hand: \(F_h = F_g = 19.62 \, \text{N}\) So, the magnitude of the total force of the man's hand on the stone is \(19.62 \, \text{N}\).

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Most popular questions from this chapter

A box full of books rests on a wooden floor. The normal force the floor exerts on the box is \(250 \mathrm{N}\). (a) You push horizontally on the box with a force of \(120 \mathrm{N},\) but it refuses to budge. What can you say about the coefficient of static friction between the box and the floor? (b) If you must push horizontally on the box with a force of at least \(150 \mathrm{N}\) to start it sliding, what is the coefficient of static friction? (c) Once the box is sliding, you only have to push with a force of \(120 \mathrm{N}\) to keep it sliding. What is the coefficient of kinetic friction?
While an elevator of mass 832 kg moves downward, the tension in the supporting cable is a constant \(7730 \mathrm{N}\) Between \(t=0\) and \(t=4.00 \mathrm{s},\) the elevator's displacement is \(5.00 \mathrm{m}\) downward. What is the elevator's speed at \(t=4.00 \mathrm{s} ?\)
Yolanda, whose mass is \(64.2 \mathrm{kg},\) is riding in an elevator that has an upward acceleration of \(2.13 \mathrm{m} / \mathrm{s}^{2} .\) What force does she exert on the floor of the elevator?
A 2010 -kg elevator moves with an upward acceleration of $1.50 \mathrm{m} / \mathrm{s}^{2} .$ What is the force exerted by the cable on the elevator?
The forces on a small airplane (mass \(1160 \mathrm{kg}\) ) in horizontal flight heading eastward are as follows: gravity \(=16.000 \mathrm{kN}\) downward, lift \(=16.000 \mathrm{kN}\) upward, thrust \(=1.800 \mathrm{kN}\) eastward, and \(\mathrm{drag}=1.400 \mathrm{kN}\) westward. At \(t=0,\) the plane's speed is \(60.0 \mathrm{m} / \mathrm{s} .\) If the forces remain constant, how far does the plane travel in the next \(60.0 \mathrm{s} ?\)
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