You have a spring attached to the bottom of a container. The spring is aligned v

Question

You have a spring attached to the bottom of a container. The spring is aligned vertically. The container is filled with water (density 1.0 x 103 kg/m3). The spring constant of the spring is 370 N/m.

a) First, you place a brick in the water, on top of the spring. The brick is a standard UK brick (10.0 cm by 20.0 cm by 5.5 cm). It weighs 2.75 kg. What is the brick%u2019s density?

b) What is the buoyant force on the block?

c)How much does the block compress the spring?

d) Next, you tie a block of oak to the spring. This block is the same volume as the aluminum block. Its density is 0.56 x 103 kg/m3. How much does the spring stretch when the block is tied to the top of it?

Explanation / Answer

(a) Mass = 2.75

Volume = 10*20*5.5 = 1100 cm^3 = 1.1*10^-3 m^3

Density = mass/volume = 2.75/1.1*10^-3 = 2500 Kg/m^3

(b)

Buyoant force = (volume of brick)*(density of water)*g = 1.1*10^-3*1000*9.8 = 10.78 N

(c)

force by spring = k*x

Apply N.L.M.

Force of spring + buyoant force = weight of brick(m*g)

k*x + 10.78 = 2.75*9.8

kx = 16.17

x = 16.17/370 = 0.0437 m = 4.37 cm

block compresses the spring 4.37 cm

(d)

force by spring = k*x

Buyoant force = (volume of brick)*(density of water)*g = 1.1*10^-3*1000*9.8 = 10.78 N

Apply N.L.M.

Force of spring + Mass of block = Buyoant force

k*x + (density)*(volume)*g = 10.78

kx = 10.78 - 6.037 = 4.74

x = 4.74/370 = 0.0128 = 1.28 cm

the spring stretch 1.28 cm

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