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A force of 10lb is required to hold a spring stretched 4 in beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length so hookies law states f(x) = kx so I have 10lbs = k(4in) k = 10/4 so I have f(x) = (10/4)(6) = 15lbs-in My text book states the answer as 15/4 ft-lb where did I got wrong?
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U have calculated k right However, ur question asks for the WORK done and not the force. Work = k * x^2
so force is f(x) = k*x and work is f(x) = k*x^(2)
yes
thanks for clarifying
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