Question

A 425 g block is released from rest at height h0 above a vertical spring with spring constant k = 460 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 15.3 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 6h0 above the spring, what would be the maximum compression of the spring?

Answer 1

a) Work done by the block on the spring
`W_(b)` = 5.38 Joules

b) Work done by the spring on the block
`W_(s)` = - 5.38 Joules

c)
`h_(0) = 1.14 m`

d) Maximum compression of the spring, x = 0.36 m

Step-by-step explanation:

Spring constant, k = 460 N/m

mass, m = 0.425 kg

compression, x= 15.3 cm = 0.153 m

a) Work done by the block on the spring,
`W_(b)`=
`0.5kx^(2)`

`W_(b)` =
`0.5*460*0.153^(2)`

`W_(b)` = 5.38 Joules

b) Work done by the spring on the ball,
`W_(s)` = - (Work done by the block on the spring,
`W_(b)`)

`W_(s)` = -
`W_(b)`

`W_(s)` = - 5.38 Joules

c) Value of
`h_(0)`

Applying the principle of energy conservation

`mgh_(0) = mgx + 0.5kx^(2)`

Inserting the appropriate values into the energy conservation equation above

`0.425*9.8*h_(0) = -(0.425*9.8*0.153) + (0.5*460*0.153^(2) )\\4.165h_(0) = 6.0213\\h_(0) = 4.75/4.165\\h_(0) = 1.14 m`

d)Maximum compression if height = 6h₀

`mg(6h_(0)) = -mgx + 0.5kx^(2)\\0.425*9.8*(6*1.14) = -(0.425*9.8x) + (0.5*460*x^(2))\\28.49 = -4.165x + 230x^(2)\\`

Find the value of x using the quadratic formula

a = 230, b = -4.165, c = -28.49

`x = \frac{-b \pm \sqrt{b^(2)-4ac } }{2a} \\x = \frac{4.165 \pm \sqrt{(-4.165)^(2)-(4*230*(-28.49) } }{2*230}\\x = (4.165 \pm 161.95)/(460) \\`

`x = (4.165 +161.95)/(460)\\` =
`0.36`

or
`x = (4.165 -161.95)/(460)\\`
`= -0.34`

x = 0.36 m is the only realistic value

Answer 2

(a) = +5.38m (b) = -5.38m (c) = 1.246m (d) = +0.3771m.

Step-by-step explanation:

Initially the spring is at equilibrium,

Work done by all forces = change in kinetic energy

Work = ∇K.E

Work = Kf -Ki =0

Since the work done = 0 since the body is at rest.

W(spring) + W(gravity) = 0

W(spring) + W(gravity) = 0

W(spring) = -W(gravity)

Work done by the block on the spring = W(block/spring)

W(block/spring) + W(spring) = 0

W(spring) = -∫kx.dx

W(spring) = ½k(X²i - X²f) ; Xi =0, Xf = 15.3cm = 0.153m

W(spring) = -½* 460 * (0.153)²

W(spring) = - 5.38NM

Work done by block on spring = + 5.38NM

(b). Workdone by spring on the block = -5.38NM.

Note: This is so because the displacement of the force is in the opposite direction to the previous one since they counter each other to maintain equilibrium.

(C). W(spring) +W(gravity) = 0

½kx² + mg(h + x) = 0

-5.83 + mg(h + 0.153) =0

5.83 = 0.425*9.8 (h + 0.153)

5.83 = 4.165(h + 0.153)

H = 1.399 - 0.153

H = 1.246m

(D).

If the release height was 6ho

H = 6* 1.246m = 7.476m

W(spring) = W(gravity)

½kx² = mg(7.476 + x)

Note: At maximum compression, the blocks would be at rest.

½Kx² = mg(h + x)

½ * 460 * x² = 0.425 * 9.8 * (7.476 + x)

230x² = 4.165 (7.476 + x)

230x² = 31.137 + 4.165x

230x² - 4.165x - 31.137 = 0

Solving the quadratic equation ( i would suggest you use formula method for easy navigation of the variables)

X = + 0.3771m or -0.3589m

But we can't have a negative compression value,

X = + 0.3771m