In the figure, a 0.27 kg block of cheese lies on the floor of a 900 kg elevator

Question

In the figure, a 0.27 kg block of cheese lies on the floor of a 900 kg elevator cab that is being pulled upward by a cable through distance d_1 = 2.1 m and then through distance d_2 = 10.6 m. (a) Through d_1, if the normal force on the block from the floor has constant magnitude F_N = 4.16 N, how much work is done on the cab by the force from the cable? (b) Through d_2, if the work done on the cab by the (constant) force from the cable is 93.64 kJ, what is the magnitude of F_N? Number Units Number Units

Explanation / Answer

a)

m = mass of block = 0.27

a = acceleration of elevator

for the block force equation is given as

Fn - mg = ma

4.16 - 0.27 x 9.8 = 0.27 a

a = 5.61 m/s2

for the elevator , force equation is given as

T - Mg - Fn = Ma

T = 900 (9.8 + 5.61) + 4.16 = 13873.2 N

work done = T d = 13873.2 x 2.1 = 29133.72 J

b)

T = W/d2 = 93640 /10.6 = 8833.96 N

force equation is given as

T - Mg - Fn = Ma

8833.96 - 900 (9.8 + 5.61) = Fn

Fn = - 5035.04 N

b)

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