Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
Answer:
Explanation:
Given that
The window height is 2m
And the window is 7.5m from the ground
Then the total height of the window from the ground is 7.5+2=9.5m
It takes the ball 0.32sec travelled pass the window.
When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')
Now using the equation of free fall during this window travels
S=ut-½gt² against motion.
S=2, g=9.81, t=0.32sec
Then,
S=u't-½gt²
2=u'×0.32-½×9.81×0.32²
2=0.32u'-0.5023
2+0.5032=0.32u'
Then, 0.32u'=2.5032
u'=2.5032/0.32
u'=7.82m/s
This is the initial velocity as the ball got the the window
Now, let analyse from the window bottom to the ground which is a distance of 7.5m
Using the equation of free fall again
v²=u²-2gH
In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,
While u is the original initial velocity from the throw of the ball
Then,
u'²=u²-2gH
7.82²=u²-2×9.81×7.5
61.146=u²-147.15
61.146+147.15=u²
Then, u²=208.296
So, u=√208.296
u=14.43m/s
The initial velocity of the ball form the throw is 14.43m/s
Answer:
Stretch can be obtained using the Elastic potential energy formula.
The expression to find the stretch (x) is 
Explanation:
Given:
Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.
To find: Elongation in the spring (x).
We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).
The formula to find EPE is given as:

Rewriting the above expression in terms of 'x', we get:

Example:
If EPE = 100 J and spring constant, k = 2 N/m.
Elongation or stretch is given as:

Therefore, the stretch in the spring is 10 m.
So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.