Question

In the figure, a 0.23 kg block of cheese lies on the floor of a 940 kg elevator cab that is being pulled upward by a cable through distance d1 = 2.5 m and then through distance d2 = 10.4 m. (a) Through d1, if the normal force on the block from the floor has constant magnitude FN = 3.65 N, how much work is done on the cab by the force from the cable? (b) Through d2, if the work done on the cab by the (constant) force from the cable is 92.49 kJ, what is the magnitude of FN?

9:12 PM Thu Feb 6 Done< > AA A assessment.education.wiley.com = Next > Send to Gradebook < Prev Next > Question 7 --/1 View Policies Current Attempt in Progress In the figure, a 0.23 kg block of cheese lies on the floor of a 940 kg elevator cab that is being pulled upward by a cable through distance d1 = 2.5 m and then through distance d2 = 10.4 m.(a) Through d1, if the normal force on the block from the floor has constant magnitude FN-3.65 N. how much work is done on the cab by the force from the cable? (b) Through d2, if the work done on the cab by the constant) force from the cable is 92.49 kJ, what is the magnitude of FN? (a) Number i Units (b) Number i Units GO Tutorial Save for Later Attempts: 0 of 5 used Submit Answer Send to Gradebook < Prev Next >

Answer 1

i Info m= 0.23 kg M = 940 kg de = 2. 5m Provided values da = 10.4m FN = 3.65 N W - 92.49 kJ Pact(A) The net upward force = from tfn - (ma Mlg . (m+ M) a from cheese alone Fw - mg - ma a Fn-ing 3.65-0.23x9.8 (a = 6.06 m/s² m+ M) (arq) - EN = (0.23 +940) (6.06+9.8) - 365 4908.39 N Spain werdene w = Front d = 14908.39 x 25 J Ws37.270 kJ > Rueton) Part (B) work done w=mah 92.49x10 a mih (moM) da (0.23 +940) X 10.4 ja = 9.45 m/s Normal force FN = ma - 0.23 x 9.45 [Fw= 2.173 N/PUKR) a = W W