A man lifts a 2.25-kg stone vertically with his hand at a constant upward velocity of 1.39 m/s. What is the magnitude of the total force of the man's hand on the stone?
Question
A man lifts a 2.25-kg stone vertically with his hand at a constant upward velocity of 1.39 m/s. What is the magnitude of the total force of the man's hand on the stone?
Solution
To solve this problem, we need to consider two forces acting on the stone: the force exerted by the man and the force of gravity acting on the stone.
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First, calculate the force of gravity on the stone. The force of gravity (weight) can be calculated using the formula F = m*g, where m is the mass of the stone and g is the acceleration due to gravity. On Earth, g is approximately 9.8 m/s². So, the force of gravity on the stone is F = 2.25 kg * 9.8 m/s² = 22.05 N.
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Since the stone is moving at a constant velocity, the net force acting on it is zero. This means the upward force exerted by the man must be equal to the downward force of gravity.
So, the magnitude of the total force of the man's hand on the stone is 22.05 N.
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