Answer:
a) v = 4.64 m / s , b) t = 0.947 s , c) t = 0.947 s
Explanation:
We will work on this exercise with vertical launch kinematics, let's start by calculating the height of the jumper in the SI system
y₀ = 5 ’(0.3048 m / 1’) + 7 ”(2.54 10-2 m / 1”) = 1.70 m
The distance they give is the height of the jump
y = 1.10 m
Let's use energy conservation
Starting point. On the floor
Em₀ = K = ½ m v²
Final point. Maximum height
Em_{f} = U = m g y
Em₀ =
½ m v² = m g y
v = √2gy
Let's calculate
v = √(2 9.8 1.10)
v = 4.64 m / s
b) Air time is the time to go up plus the time to go down, which is the same
For maximum height the speed is zero
v = v₀ - g t₁
t₁ = v₀ / g
t₁ = 4.64 /9.8
t₁ = 0.4735 s
The total time is
t = 2 t₁
t = 2 0.4735
t = 0.947 s
c) if it takes a distance of 0.40 to reach speed, what is the acceleration, as it stands on the floor its initial speed is zero
v² = v₀² + 2 a x
a = v² / 2x
a = 4.64²/2 0.40
a = 26.9 m / s²
Answer:
a)
b)
Explanation:
Part a
For this case we can begin finding the period like this:
Then we know that the centripetal acceleration is given by:
And the velocity is given by:
If we replace this into the acceleration we got:
And we can replace the values and we got:
Part b
For this case we want to find a value of k such that:
Where a = 9.74, so then we can solve for k like this:
Answer:
W = 290.7 dynes*cm
Explanation:
d = 1/5 cm = 0.2 cm
The force is in function of the depth x:
F(x) = 1000 * (1 + 2*x)^2
We can expand that as:
F(x) = 1000 * (1 + 4*x + 4x^2)
F(x) = 1000 + 4000*x + 4000*x^2
Work is defined as
W = F * d
Since we have non constant force we integrate
W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2
W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3
W = 200 + 80 + 10.7 = 290.7 dynes*cm