How much total work is done by the force in lifting the elevator from 0.0 m to 9.0 m?
1 answer:
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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:
T=7.4 N hence T<30 N
Explanation:
The figure is likely to be similar to the one attached. Writing the equation for forces we have
F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as
30 N - T = (30/9.81)a
Also, we know that T=F*a/g and substituting 10N for F we obtain the second equation as
T = (10/9.81)a
Adding the first and second equations we obtain
30 = 4.077471967
a Hence
and T=a hence
T is approximately 7.4 N
