<span> force of 10.0 N
</span>
<span>distance of 0.9 m
w=f*d
w=10*0.9
=9.0 j</span>
If the collision is elastic, there is no loss in kinetic energies, which means that the total energies before and after impact are the same. So no need to worry about final velocities.
Final energy
= initial energy
= (1/2) (7.20*2.00^2+5.75*(-1.30)^2)
=19.26 joules
Answer: the total kinetic energy is 19.3 J. after collision.
Answer:
m1=914.9kg
m2=604.9kg
m3=864.75kg
Explanation
I think we are suppose to find the mass of the crate.
The effective force that moves the body in positive x direction is 3615N
ΣFx = Σma
Then Fx=3615N
Then the masses be m1, m2 and m3
Then,
ΣF = Σ(ma)
3615=(m1+m2+m3)a
Given that a=1.516
The masses are
m1+m2+m3=, 2384.56. Equation 1
Between mass 1 and mass 2 is, F12=1387.
The effective force that pull mass 1 is 1387.
F12=m1 ×a
Therefore,
m1=F12/a
m1=1387/1.516
m1=914.9kg.
The effective force that pulls crate 1 and crate 2 is F23
F23=(m1+m2)a
Therefore
2304=(m1+m2)a
Therefore, since a=1.516
m1+m2=2304/1.516
m1+m2=1519.8kg
Since m1=914.9kg
So, m2=1519.8-m1
m2=1519.8-914.9
m2=604.9kg
Also from equation 1
m1+m2+m3=2384.56
Since m1=914.9kg and m2=604.9kg
Then, m3=2384.56-604.9-914.9
m3=864.75kg
Answer:
Joanna's position would be -100.
Explanation:
Joanna's position would be 100 feets west to her house. Her position is negative because Joanna walked west of her house. The position would have been positive if she had moved east of his house, that is, to the school.