A 89-kg safe is dropped vertically from rest form the top of a building 26 m abo

Question

A 89-kg safe is dropped vertically from rest form the top of a building 26 m above a huge spring (spring constant k = 4.1x104 N/m) at the bottom of the building on the ground. Safe finally comes to rest by compressing the spring. Assume that y = 0 is at the equilibrium position of the a) b) c) d) e) f) spring. Calculate the work done by gravity on the safe during the fall. Calculate the speed of the safe just before striking the spring. Calculate the amount (in meter) the spring compresses when the safe lands on it (note that work is done by both spring and gravity in this part). Calculate the elastic potential energy of the compressed spring. Calculate the potential energy of the safe after it landed on the spring. Calculate the deceleration of the safe when it compresses the spring.

Explanation / Answer

m =89kg , h =26 m, k =4.1x10^4 N/m

(a) W =mgh =89*9.8*26 = 22677.2 J

(b) From conservation of energy

Kf =Ui

(1/2)mv^2 =mgh

v =[2gh]1/2 =[2*9.8*26]1/2

v = 22.6 m/s

(c) W = mgh = (1/2)kx2

x = [2W/k]1/2 =[(2*22677.2)/(4.1x104)]1/2

x =1.05 m

(d) U =(1/2)kx2= 22677.2 J

(e) U =mgx = 89*9.8*1.05 = 915.81 J

(f) mgx =(1/2)mvf2

vf = [2gx]1/2 =[2*9.8*1.05]1/2

vf = 4.54 m/s

vf^2 -v^2 =2ax

(4.5)2 -(22.6)2= 2a(1.05)

a = - 233.6 m/s2

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