The algebraic expression for the component of the normal force in the vertical direction is Force= product of mass and area.
<h3>
What is Force?</h3>
A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction. Or, there is a specific meaning to the word "force." At this level, calling a force a push or a pull is entirely appropriate. A force is not something an object "has in it" or that it "contains." One thing experiences a force from another. There are both living things and non-living objects in the concept of a force.
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Answer:
u" + 40u' + 49u = 2 sin(t/6)
upp + 40up + 49u = 2 sin(t/6)
Explanation:
Step 1: Data given
mass = 5 kg
L = 20 cm = 0.2 m
F = 10 sin(t/6)N
Fd(t) = - 6 N
u(0) = 0.03 m/s
u(0) = 0
u'(0) = 3 cm/s
Step 2:
ω =kL
k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²
Since Fd(t) = -γu'(t) we know:
γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m
The initial value problem which describes the motion of the mass is given by
5u" + 200u' + 245u = 10 sin(t/6) u(0) = 0 ; u'(0) = 0.03
This is equivalent to:
u" + 40u' + 49u = 2 sin(t/6) u(0) = 0 ; u'(0) = 0.03
upp + 40up + 49u = 2 sin(t/6)
With u in m and t in s
The frequency will not change
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
