WILL GIVE BRAINLIEST AND 100 POINTS! NEED THIS ASAP!
1 answer:
You might be interested in
First, let's find the speed 
of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
After the collision, the two blocks stick together and so now they have mass
and they are moving with speed :
For conservation of momentum
So we can write
From which we find
The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force
. The work done by the frictional force to stop the two blocks is
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
From which we can find the value of the coefficient of kinetic friction:
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
After the collision, the two blocks stick together and so now they have mass
For conservation of momentum
So we can write
From which we find
The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
From which we can find the value of the coefficient of kinetic friction:
Answer:
The energy of this particle in the ground state is E₁=1.5 eV.
Explanation:
The energy of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:
In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:
So we can rewrite the energy in the ground state as:
Finally