A man lifts a 1.79-kg stone vertically with his hand at a constant upward acceleration𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 of 1.46 m/s22. What is the magnitude of the total force of the man's hand on the stone?
Question
A man lifts a 1.79-kg stone vertically with his hand at a constant upward acceleration𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 of 1.46 m/s22. What is the magnitude of the total force of the man's hand on the stone?
Solution
To solve this problem, we need to use Newton's second law of motion, which states that the force acting on an object is equal to its mass times its acceleration (F = ma).
Step 1: Identify the given values. The mass (m) of the stone is 1.79 kg and the acceleration (a) is 1.46 m/s².
Step 2: Calculate the force due to gravity. The force due to gravity (Fg) on the stone can be calculated using the formula Fg = m*g, where g is the acceleration due to gravity. On Earth, g is approximately 9.8 m/s². So, Fg = 1.79 kg * 9.8 m/s² = 17.542 N.
Step 3: Calculate the force due to the man's hand. The force due to the man's hand (Fh) can be calculated using the formula Fh = m*a. So, Fh = 1.79 kg * 1.46 m/s² = 2.614 N.
Step 4: Calculate the total force. The total force (Ft) is the sum of the force due to gravity and the force due to the man's hand. So, Ft = Fg + Fh = 17.542 N + 2.614 N = 20.156 N.
Therefore, the magnitude of the total force of the man's hand on the stone is 20.156 N.
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