Respuesta :
Explanation:
Given that,
Mass of the person, m = 72.5 kg
Initial speed of the car, u = 20.5 m/s
(a) Let [tex]F_x[/tex] is the horizontal component of the average force, in newtons, on the person if he is stopped by a padded dashboard that compresses an average of 1.15 cm, x = 0.0115 m and v = 0
Using the second law of motion to find it as:
[tex]F_x=m\times a[/tex]
Using the third equation of kinematics to find a.
[tex]F_x=m\times \dfrac{v^2-u^2}{2s}[/tex]
[tex]F_x=-m\times \dfrac{u^2}{2s}[/tex]
[tex]F_x=-72.5\times \dfrac{(20.5)^2}{2\times 0.0115}[/tex]
[tex]F_x=-1.32\times 10^6\ N[/tex]
(b) Here, x = 13.5 cm = 0.135 m
[tex]F_x=-m\times \dfrac{u^2}{2s}[/tex]
[tex]F_x=-72.5\times \dfrac{(20.5)^2}{2\times 0.135}[/tex]
[tex]F_x=-1.12\times 10^5\ N[/tex]
Hence, this is the required solution.
Answer:
A) f_{dashboard} = 1324701 N
B) f_{airbag} = 112844.90 N
Explanation:
Given data:
mass of person = 72.5 kg
velocity of person = 20.5 m/s
we know that from newton's 2nd law
f = ma
[tex]f = m \frac{v^2 - u^2}{2s}[/tex]
final velocity is v = 0
[tex]f = - \frac{mu^2}{2s}[/tex]
[tex]f = \frac{72.5\times 20.5^2}{2\times 1.15\times 10^{-2}}[/tex]
f = 1324701 N
B)
[tex] f_{airbag} = f = \frac{72.5\times 20.5^2}{2\times 1.3.5\times 10^{-2}}[/tex]
[tex] f_{airbag} = 112844.90 N[/tex]
