A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is sud

Question

3 years ago

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A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is sud

denly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

Engineering

2 answers:

Inessa05 [86] · Answer 1 · 2022-01-01T20:47:14+03:00

<h2></h2><h2>Answer:</h2><h2></h2>

(a) 5.59m/s

(b) 1.56m

<h2>Explanation:</h2>

<em>Analysis 1:</em> When the 3-kg block rests on the 2-kg block, their total weight, W is given as;

W = M x g -------------------(i)

Where;

M = total mass = 3kg + 2 kg = 5kg

g = acceleration due to gravity = 10m/s²

<em>Substitute these values into equation (i) as follows;</em>

W = 5 x 10 = 50N

<em>Analysis 2: </em>This weight, W, will cause a compression, c, on the spring and this is given according to Hooke's law by;

F = k x c ---------------(ii)

Where;

F = force acting on the spring = W = 50N

k = the spring's constant = 40N/m

c = compression on the spring

<em>Substitute these values into equation (ii) as follows;</em>

50 = 40 x c

c = $\frac{50}{40}$

c = 1.25m

Therefore, the compression caused by the load (the masses) is 1.25m

<em>Analysis 3: </em>This compression will cause an elastic energy ( $E_{E}$ ) in the spring which is given by Hooke's law as follows;

$E_{E}$ = $\frac{1}{2}$ x k x c² ----------------(iii)

Substitute the values of k and c into equation (iii) as follows;

$E_{E}$ = $\frac{1}{2}$ x 40 x 1.25²

$E_{E}$ = 31.25J

Therefore, the elastic energy produced by the spring is 31.25J

(a) Now, when the upper block (the 3-kg mass) is removed, the 2-kg block moves with some initial velocity (v) thereby resulting into kinetic energy by the 2-kg block. In other words, the elastic energy ( $E_{E}$ ) in the spring is converted into kinetic energy ( $K_{E}$ ) by the block. i.e

$E_{E}$ = $K_{E}$ -------------------------(iv)

Where;

$K_{E}$ = $\frac{1}{2}$ x m x v² [m = mass of the 2-kg block]

<em>Substitute the value of </em> $K_{E}$ <em> into equation (iv) as follows;</em>

$E_{E}$ = $\frac{1}{2}$ x m x v²

Where;

$E_{E}$ = 31.25J

m = 2kg

<em>Substitute these values into equation(v) as follows;</em>

31.25 = $\frac{1}{2}$ x 2 x v²

v² = 31.25

v = $\sqrt{31.25}$

v = 5.59m/s

Therefore, the initial velocity is 5.59m/s. This is also the maximum velocity reached by the block since velocity decreases with elevation such that at maximum height, velocity is zero.

(b) At maximum height, the elastic energy is fully converted to potential energy ( $P_{E}$ ). i.e