Question

A mass M is on a horizontal, frictionless surface, and is initially at rest. It is subject to a position dependent force F =kin'i. If k=62.6N, y=1.2, the mass is initially at ři=1.2mi, how much work is done as it move along the x-axis to a final position 5.5mi? Answer in J. Answer: 1550 * The correct answer is: 1170 Try another question like this one

Answer 1

Given force F= K (njo î The displacement is along the x-axis, the infinitesimal displacement is 2 di= da The infinitesimal work done in displacing the mass by an infinitesimal amount is dW= F. dr dW=k( x ) ho de Using î.îal dWok (x) dx The mass is initially at x, = 1,2m, the final position is a = 5.5m. To find the total work done by force we integrate dw between these two points W = dwo k (x te lam (Imus Note that 8 = 1.2= 12 = 8 is a rational number. So we can use Sx dx = a + Const. com 5.5. x=5.5m at a+1 at

5 m Xat W= 0,453 .602" [15.5m)* – 12m3] W = (62.6 %), 41(m) 22 am)''2 22^ W = 1167 Nm In three significant IW=1170 J [L Nm = 13]