Question

In the figure, a 0.23 kg block of cheese lies on the floor of a 970 kg elevator cab that is being pulled upward by a cable through distance d1 = 2.1 m and then through distance d2 = 11.5 m. (a) Through d1, if the normal force on the block from the floor has constant magnitude FN = 3.60 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 91.60 kJ, what is the magnitude of FN?

Answer 1

elevator cab = 970 kg

cheese block = 0.23 kg

elevator cab pulled by cable through d1 = 2.1 m

Normal force on block = 3.60 N

weight of block = mg = 0.23*9.8 = 2.25 N

weight of the block acts downward

cab floor reaction pushes the block upward.

net force on the block F= 3.60 - 2.25 = 1.35 N

acceleration of the block a = F/m = 1.35 /0.23 = 5.87 m/s/s

The block and the cab move with the same direction as the block is in contact with the cab floor.

Ma = T-Mg

Tension in the cable T = M(a+g) = 970(9.8+5.87)

work done by the cable force = T*d₁ = 970(9.8+5.87)*2.1 = 31919 J

b) d2 = 11.5 m

work done by the cable = 91.6 kJ

cable tension = T = 91.6E+3/11.5 = 7965.2 N

cab acceleration a = 7965.2/970 - 9.8 = -1.59 m/s/s

The cab is moving down, hence a is -ve, Normal reaction on the block would be less than its weight.

F_N = m(g-a) = 0.23(9.8-1.59) = 1.70 N