The thermal efficiency is defined as follows
,
and the energy which is put into the system is
.
In your case
So
which gives an efficiency of
.
Net force can be defined as the sum of force acting on an object in all the directions
.i.e Net force= F acting on a body at upward direction+ F at downward direction+ F at leftward direction+ F at rightward direction.
e.g if a body is at rest the sum of forces acting on body in all direction is zero
but if a body is moving its mean body is facing an unbalanced net force.
Answer:
a) x=63.0 
or 6.3cm
b) x=116.0 or 11.6cm
Explanation:
a).
The elastic potential energy is modeling by equation :
b).
The work energy theorem explain which work is done in this case. the motion began from the rest so K1=K2 equal zero, Ug1 is no yet done and U2is also zero because is the potential energy
Solving for x
The negative is discard so
x=0.116m
The answer is not 'A'.
Using the numbers given in the question, it's 'B'.
It WOULD be 'A' if the number in B were 71 or less.
________________________________
<span>This is not as simple as it looks.
What quantity are we going to compare between the two cases ?
Yes, I know ... the "amount of work". But how to find that from the
numbers given in the question ?
Is it the same as the change in speed ?
Well ? Is it ?
NO. IT's NOT.
In order to reduce the car's speed, the brakes have to absorb
the KINETIC ENERGY, and THAT changes in proportion to
the SQUARE of the speed. ( KE = 1/2 m V² )
Case 'A' :
The car initially has (1/2 m) (100²)
= (1/2m) x 10,000 units of KE.
It slows down to (1/2 m) x (70²)
= (1/2m) x 4,900 units of KE.
The brakes have absorbed (10,000 - 4,900) = 5,100 units of KE.
Case 'B' :
The car initially has (1/2 m) (79²)
= (1/2m) x 6,241 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 6,241 units of KE.
Just as we suspected when we first read the problem,
the brakes do more work in Case-B, bringing the car
to a stop from 79, than they do when slowing the car
from 100 to 70 .
But when we first read the problem and formed that
snap impression, we did it for the wrong reason.
Here, I'll demonstrate:
Change Case-B. Make it "from 71 km/h to a stop".
Here's the new change in kinetic energy for Case-B:
The car initially has (1/2 m) (71²)
= (1/2m) x 5,041 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 5,041 units of KE.
-- To slow from 100 to 70, the brakes absorbed 5,100 units of KE.
-- Then, to slow the whole rest of the way from 71 to a stop,
the brakes absorbed only 5,041 units of KE.
-- The brakes did more work to slow the car the first 30 km/hr
than to slow it to a complete stop from 71 km/hr or less.
That's why you can't just say that the bigger change in speed
requires the greater amount of work.
______________________________________
It works exactly the same in the opposite direction, too.
It takes less energy from the engine to accelerate the car
from rest to 70 km/hr than it takes to accelerate it the
next 30, to 100 km/hr !</span>
(The exact break-even speed for this problem is 50√2 km/h,
or 70.711... km/hr rounded. )
