Answer:
(a). The spring compressed is
.
(b). The acceleration is 1.5 g.
Explanation:
Given that,
Acceleration = a
mass = m
spring constant = k
(a). We need to calculate the spring compressed
Using balance equation

....(I)
The spring compressed is
.
(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,
The compression is given by

Here, acceleration is zero
So, 
We need to calculate the acceleration
Put the value of x in equation (I)




Hence, (a). The spring compressed is
.
(b). The acceleration is 1.5 g.
Answer:


Explanation:

Required
The units of A and B
From the question, we understand that:

This implies that each of
and
will have the same unit as 
So, we have:


The unit of t is (s); So, the expression becomes

Divide both sides by 


Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
El beneficio que obtuvo es de 2 L