Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer : I hope this helps !
The Effort Force is the force applied to a machine. Work input is the work done on a machine. The work input of a machine is equal to the effort force times the distance over which the effort force is exerted.
Answer:
Its not A..
Explanation:
I chose A - was incorrect
Answer:
Calculating Coefficient of friction is 0.229.
Force is 4.5 N that keep the block moving at a constant speed.
Explanation:
We know that speed expression is as
.
Where,
is initial speed, V is final speed, ∆s displacement and a acceleration.
Given that,
=3 m/s, V = 0 m/s, and ∆s = 2 m
Substitute the values in the above formula,

0 = 9 - 4a
4a = 9

is the acceleration.
Calculating Coefficient of friction:


Compare the above equation

Cancel "m" common term in both L.H.S and R.H.S





Hence coefficient of friction is 0.229.
calculating force:


F = 4.5 N
Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.
Answer:
The hollow cylinder rolled up the inclined plane by 1.91 m
Explanation:
From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

moment of inertia, I, of a hollow cylinder = ¹/₂mr²
substitute for I in the equation above;


given;
v₁ = 5.0 m/s
vf = 0
g = 9.8 m/s²

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m