Question

A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal frictionless surface once the spring is stretched and released. A 1.00-kg block of wood rests on top of the first block. The coefficient of static friction between the two blocks of wood is 0.450. What is the maximum speed that this set of blocks can have as it oscillates if the top block of wood is not to slip?

Answer 1

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

$\omega=\sqrt{\dfrac{k}{M+m}}$

By putting the values

$\omega=\sqrt{\dfrac{150}{5.67+1}}$

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

$\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2$

$kX^2=(m+M)v^2$

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

Answer 2

Answer:

V = 0.9294 ms-1

Explanation:

The following data is provided in the question:

Mass of first block = m1 = 5.67 kg  

Mass of 2nd block = m2 = 1 Kg

Spring constant = K = 150 N/m

Coefficient of static friction = µ = 0.450

The acceleration for 2nd block will be  

a = µg = (0.450)(9.8)

a = 4.41 m/s^2       .…………. (1)

In simple harmonic motion the acceleration is given as:  

a = (ω^2)X                …………… (2)

where, ω is natural frequency and X is the displacement.

We know that:

ω =√(K/(m1+m2)) = (150/6.67)^0.5

ω = 4.74 rad/s   …………… (3)

Put equation (1) and (3) in (2) and solving for X,

X = 0.196 m

As the equation for energy conversion is:

½ KX2 = ½ (m1 + m2)V^2   …………. (4)

By putting the values and solving for V,

V = 0.9294 ms-1