Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
Is this a true and false question?
Answer:
Given that
m = 5.3 kg
Fx = 2x + 4
We know that work done by force F given as
w= ∫ F. dx
a)
Given that x=1.08 m to x=6.5 m
Fx = 2x + 4
w= ∫ F. dx
![w=\left [x^2+4x \right ]_{1.08}^{6.5}](https://tex.z-dn.net/?f=w%3D%5Cleft%20%5Bx%5E2%2B4x%20%5Cright%20%5D_%7B1.08%7D%5E%7B6.5%7D)
w=62.7 J
b)
We know that potential energy given as
∫ dU = -∫F.dx ( w= ∫ F. dx)
ΔU= -62.7 J
c)
We know that form work power energy theorem
Net work = Change in kinetic energy
W= KE₂ - KE₁
62.7 =KE₂ - (1/2)x 5.3 x 3²
KE₂ = 86.55 J
This is the kinetic energy at 6.5m
