A force of 10 lb 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?
Applications of Integration
given that a force of £10 is required to hold the spring stretched four inch beyond its natural land. How much work is done? And stretching it from its natural length, natural length to six inches beyond its natural end. So first of let's use this information to find the spring constant. So we know that force is equal to K. X. Not were ex not as the lent by which it has been stressed or compressed. So in this case 10 is equal to K times X. Not it's not has given us for So the value of cake um sort of stand over four which is nothing but fire over to and this is pounds per inch. Uh Now we can use the formula for the for the world done from from its natural and to six inches beyond its natural. And so the work done is the change in the potential energy of the spring. So that's the potential energy. Final minus the potential energy issue. Final protection energy is going by K over two X. North square. The final the amount of extension minus initial potential energy is zero. Because the desert is natural. Then K. Is already given. We founded as five over to over to remains other tests. And and that stretches six inches. About six square That will come as 5/4 times 36. Four names 1936. And this is 45. So 45. And the unit won't really be jewel that we have to take care about that. And what are the unit is So since the forces and pounds and the displacement is a is an inches, so pound times inch, that's going to be the unit of this particular walk down. So this is the required answer. Thank you.