when a 2.00-kg object is hung vertically on a certain light spring described by hooke's law, the spring stretches 2.26 cm. (a) what is the force constant of the spring? 8.67 your response is off by a multiple of ten. n/m (b) if the 2.00-kg object is removed, how far will the spring stretch if a 1.00-kg block is hung on it? 1.13 cm (c) how much work must an external agent do to stretch the same spring 8.50 cm from its unstretched position?

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16-12-2022
Physics

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when a 2.00-kg object is hung vertically on a certain light spring described by hooke's law, the spring stretches 2.26 cm. (a) what is the force constant of the spring? 8.67 your response is off by a multiple of ten. n/m (b) if the 2.00-kg object is removed, how far will the spring stretch if a 1.00-kg block is hung on it? 1.13 cm (c) how much work must an external agent do to stretch the same spring 8.50 cm from its unstretched position?

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qwterabite qwterabite · Answer 1 · 2022-12-22T03:21:06-05:00

(a) The force constant of the spring is 8.67 N/m

(b)If the 2.00-kg object is removed, the spring will stretch 1.13 cm when a 1.00-kg block is hung on it.

(c) In order to stretch the same spring 8.50 cm from its unstretched position, an external agent must do 10.68 J of work.

The force constant of the spring, or spring constant, is the amount of force (in newtons) required to stretch the spring a certain distance (in meters). In this case, the spring stretches 2.26 cm when a 2.00-kg object is hung on it, so the force constant must be 8.67 N/m. This is calculated by dividing the mass of the object (2.00 kg) by the distance the spring stretched (2.26 cm), and then multiplying by 1000 to convert the result from kg/cm to N/m.If the 2.00-kg object is removed, the spring will stretch 1.13 cm when a 1.00-kg block is hung on it. This can be calculated by dividing the mass of the object (1.00 kg) by the distance the spring stretched when the 2.00 kg object was hung on it (2.26 cm), and then multiplying by 1000 to convert the result from kg/cm to N/m. The resulting number (1.13 cm) is how far the spring will stretch when the 1.00-kg block is hung on it. In order to stretch the same spring 8.50 cm from its unstretched position, an external agent must do 10.68 J of work. This can be calculated by multiplying the force constant (8.67 N/m) by the distance the spring is being stretched (8.50 cm) and then multiplying by 1000 to convert the result from N/cm to J. The resulting number (10.68 J) is the amount of work that must be done to stretch the spring 8.50 cm.

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