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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) If a spacecraft moves in a straight line between the Earth and the Sun, at what point would the force of gravity on the spacecraft due to the Sun be as large as that due to the Earth? (b) If the spacecraft is close to, but not at, this equilibrium point, does the net force on the spacecraft tend to push it toward or away from the equilibrium point? [Hint: Imagine the spacecraft a small distance \(d\) closer to the Earth and find out which gravitational force is stronger.]
The coefficient of static friction between a block and a horizontal floor is \(0.40,\) while the coefficient of kinetic friction is \(0.15 .\) The mass of the block is \(5.0 \mathrm{kg} .\) A horizontal force is applied to the block and slowly increased. (a) What is the value of the applied horizontal force at the instant that the block starts to slide? (b) What is the net force on the block after it starts to slide?
The coefficient of static friction between a brick and a wooden board is 0.40 and the coefficient of kinetic friction between the brick and board is $0.30 .$ You place the brick on the board and slowly lift one end of the board off the ground until the brick starts to slide down the board. (a) What angle does the board make with the ground when the brick starts to slide? (b) What is the acceleration of the brick as it slides down the board?
Which of the fundamental forces has the shortest range, yet is responsible for producing the sunlight that reaches Earth?
Which of the fundamental forces governs the motion of planets in the solar system? Is this the strongest or the weakest of the fundamental forces? Explain.
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