What is the difference between the law of conservation of energy and conserving energy
1 answer:
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Answer with Explanation:
We are given that
Initial velocity,u=4.5 m/s
Time=t =0.5 s
Final velocity=v=0m/s
We have to find the deceleration and estimate the force exerted by wall on you.
We know that
Acceleration=
Using the formula
Acceleration=
deceleration=a=
We know that
Force =ma
Using the formula and suppose mass of my body=m=40 kg
The force exerted by wall on you
Force=
Answer:
b) k Δx - W cos θ - μ mg cos θ = m a , c) θ = 86.6º, d) Δx = 1.18 m
Explanation:
a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.
F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring
b) Let's apply Newton's second law for when the spring is compressed
let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Y axis
N - W_y = 0
N = W_y
N = W cos θ
X axis
F -Wₓ -fr = ma
the force applied by the spring is given by hooke's law
F = k Δx
friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
k Δx - W cos θ - μ mg cos θ = m a ( 1)
c) If the plane has no friction, what is the angle so that Δx = 0.1m
We write the equation 1, with fr = 0 and since the system is still a = 0
k Δx - W cos θ -0 = 0
cos θ =
cos θ =
cos θ = 0.0598
θ = cos⁻¹ 0.0598
θ = 86.6º
d) In this part they give the angle θ = 45º and there is no friction, they ask the compression
the acceleration is zero, we substitute in 1
k Δx - W cos θ - 0 = 0
Δx =
Δx =
Δx = 1.18 m
A string with linear density 0.500 g/m.
Tension 20.0 N.
The maximum speed
The energy contained in a section of string 3.00 m long as a function of .
We are given following data for string with linear density held under tension :
μ = 0.5
= 0.5 x 10⁻³
T = 20 N
If string is L = 3m long, total energy as a function of is given by:
E = 1/2 x μ x L x ω² x A²
= 1/2 x μ x L x
= 7.5 x 10⁻⁴
So, The total energy as a function of = 7.5 x 10⁻⁴
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