A block weighing 40 kg extends a spring by 0.16m from its unscratched position. What is the value of k?
Correct answer: B. 245 kg/s2
- A. 170 kg/s2
- B. 245 kg/s2
- C. 215 kg/s2
- D. 201 kg/s
Explanation
Hooke's Law describes the relationship between the force exerted by a spring (F) and its displacement (x) from the equilibrium position. The formula is given by:F=−kxWhere F is the force exerted by the spring,k is the spring constant,x is the displacement from the equilibrium position.The negative sign indicates that the force exerted by the spring is in the opposite direction to the displacement.Given that the block weighs 40kg and the spring is extended by 0.16m, the force exerted by the spring (F) can be calculated using the weight (W=mg), where m is the mass and g is the acceleration due to gravity:F= mgmg=-kxk= -2450N/mSince the spring constant (k) is a positive value (magnitude), the answer is:k= 2450 N/m
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