An elastic constant spring k = 620N/m is held in a vertical position, with the lower end supported by a horizontal surface.

Question

The upper extremity is compressed 25 cm, and a block with a weight of 50N is placed on the spring, and the system is released. Assuming that the potential gravitational energy Ug of the block is zero at the point (y = 0) at which the system is released, determine the kinetic energy K of the block to y equal to:

(a) 0m,

(b) 0; 050m,

(c) 0; 10m,

(d) 0; 15m,

(e) 0; 20m.

(f) Also calculate the value of y for the maximum height reached by the block.

Answer 1

Draw a picture:

The total energy (E_TOTAL) at t = 0 (compressed spring) is the sum of the spring potential energy (PE_SPRING), the gravitational potential energy (PE_GRAVITY), and the kinetic energy (KE) or:

E_TOTAL = PE_SPRING + PE_GRAVITY + KE

We also know that:

PE_SPRING = 1/2kx²

PE_GRAVITY = mgh or in this case mgy = Wy (because weight = mg)

KE = 1/2mv² or, in this case (since W = mg, or m = W/g) = 1/2(W/g)v²

I will assume g = 9.8 m/s²

I will also assume there are 2 sig figs in the problem even though there is actually 1 (W = 50 N). If you want to adjust this to 1 sig fig, go right ahead.

At t = 0, y is also equal to 0 and all the energy is PE_SPRING because:

KE = 0 since v = 0 [there is, instantaneously, no velocity yet (you just that instant let it go so it hasn't gone anywhere yet)]

PE_GRAVITY = 0 because y = 0 and PE_GRAVITY = Wy

So E_TOTAL = PE_SPRING + PE_GRAVITY + KE = 1/2kx² + 0 + 0 = 1/2(620)(0.25)² = 19.375 joules

Since this is the total energy and since energy is conserved, no matter what instance you select, the total energy will always be 19.375 joules.

When y = 0.05 m, the spring is now partially compressed (was 0.25, now 0.20):

E_TOTAL = 19.375 joules = PE_SPRING + PE_GRAVITY + KE = 1/2kx² + Wy + 1/2(W/g)v² or

19.375 = 1/2(620)(0.20)² + (50)(.05) + 1/2(50/9.8)v² or

19.375 = 12.4 + 2.5 + 2.551v²

4.475 = 2.551v²

1.7542 = v²

v = 1.3 m/s

When y = 0.1 m, the spring is now less compressed (was 0.25, now 0.15)

E_TOTAL = 19.375 joules = PE_SPRING + PE_GRAVITY + KE = 1/2kx² + Wy + 1/2(W/g)v² or

19.375 = 1/2(620)(0.15)² + (50)(.10) + 1/2(50/9.8)v² or

19.375 = 6.975 + 5 + 2.551v²

7.4 = 2.551v²

2.9008 = v²

v = 1.7 m/s

When y = 0.15 m, the spring is now less compressed (was 0.25, now 0.10)

E_TOTAL = 19.375 joules = PE_SPRING + PE_GRAVITY + KE = 1/2kx² + Wy + 1/2(W/g)v² or

19.375 = 1/2(620)(0.10)² + (50)(.15) + 1/2(50/9.8)v² or

19.375 = 3.1 + 7.5 + 2.551v²

8.775 = 2.551v²

3.4398 = v²

v = 1.9 m/s

When y = 0.20 m, the spring is now less compressed (was 0.25, now 0.05)

E_TOTAL = 19.375 joules = PE_SPRING + PE_GRAVITY + KE = 1/2kx² + Wy + 1/2(W/g)v² or

19.375 = 1/2(620)(0.05)² + (50)(.20) + 1/2(50/9.8)v² or

19.375 = 0.775 + 10 + 2.551v²

8.60 = 2.551v²

3.3712 = v²

v = 1.8 m/s

To find the maximum height we would WANT to assume the spring is fully decompressed and the block has reached some y value above the spring where the velocity is zero. That would mean PE_SPRING = 0 and KE = 0 and we could solve for y. However, we don't really know that the spring fully decompresses. So let's check that first. At y = 0.25 (spring is fully decompressed) we have:

E_TOTAL = 19.375 joules = PE_SPRING + PE_GRAVITY + KE = 1/2kx² + Wy + 1/2(W/g)v² or

19.375 = 1/2(620)(0)² + (50)(.25) + 1/2(50/9.8)v² or

19.375 = 0 + 12.5 + 2.551v²

6.875 = 2.551v²

2.695 = v²

v = 1.6 m/s so there is still kinetic energy meaning the block is still rising. So now we can use PE_SPRING = 0 and KE = 0 to solve for the maximum height:

E_TOTAL = 19.375 joules = PE_SPRING + PE_GRAVITY + KE = 0 + Wy + 0 or

19.375 = 0 + 50y + 0

19.375 = 50y

y = 0.3875 m or y = 0.39 m