Answer:
W = 270.9 J
Explanation:
given,
F(x) = (12.9 N/m²) x²
work = Force x displacement
dW = F .dx
the push-rod moves from x₁= 1 m to x₂ = 4 m
integrating the above



![W = 12.9\times [\dfrac{x^3}{3}]_1^4 dx](https://tex.z-dn.net/?f=W%20%3D%2012.9%5Ctimes%20%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%5D_1%5E4%20dx)
`![W = 12.9\times [\dfrac{4^3}{3}-\dfrac{1^3}{3}] dx](https://tex.z-dn.net/?f=W%20%3D%2012.9%5Ctimes%20%5B%5Cdfrac%7B4%5E3%7D%7B3%7D-%5Cdfrac%7B1%5E3%7D%7B3%7D%5D%20dx)
W = 270.9 J
work done by the motor is W = 270.9 J
(a) 
First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:

where
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 1.00 cm = 0.01 m is the displacement of the person
Solving for a,

And the average force on the person is given by

with m = 75.0 kg being the mass of the person. Substituting,

where the negative sign means the force is opposite to the direction of motion of the person.
b) 
In this case,
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 15.00 cm = 0.15 m is the displacement of the person with the air bag
So the acceleration is

So the average force on the person is

When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
<h3>
Frictional force between the block and the horizontal surface</h3>
The frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Learn more about frictional force here: brainly.com/question/4618599