There are two photos posted. Part (a) to (e) are correct answers, but I couldnt

Question

There are two photos posted. Part (a) to (e) are correct answers, but I couldnt figure out part (f) and (g). WRONG ANSWERS -0.7082988J and 0.7082988J for part (f). -0.2943J and 0.2943J for part (g). Please give the right answers and show work. Thank you.

-/7 points You have two equal masses m1 and m2 and a spring with a spring constant k. The mass m is connected to the spring and placed on a frictionless horizontal surface at the relaxed position of the spring. You then h horizontal ang mass m2, connected to mass m, by a massless cord, over a pulley at the edge of the surface. When the entire system comes to rest in the equilibrium position, the spring is stretched an amount di as s hown in figure (a). You are given the following information. ?F- The mass m1 = m2 0.500 kg. The spring constant k 230 N/m. ? >

Explanation / Answer

g) m2 is connected to the rest of the system only by the string. So any work done on m2 will have to be done by the string. Work done on the system by the spring is given as 0.7082988. This work can be divided between m1 and m2 in the ratio of their masses = 0.7082988/2 = 0.3541494 Joules, which is the work done by the tension on m2.

f) Work done by gravity is entirely on m2. So net work done on m2 = 0.3541494 - 0.2943 = 0.0598J

F.S = maS= 0.0598

a = 0.0598/(0.5 x 0.06) = 1.9933 m/s2

Work done by tension on m1 = work done by spring on m1 - m.a.s = 0.3541494 - (0.5 x 1.9933 x 0.06) = 0.2943J

Since direction of force and motion are opposing, work done by tension on m1 = -0.2943J

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