Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
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The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.
Answer: Option B
<u>Explanation:
</u>
Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.
Answer:
Energy lost due to friction is 22 J
Explanation:
Mass of the ball m = 4 kg
Initially velocity of ball v = 6 m/sec
So kinetic energy of the ball 

Now due to friction velocity decreases to 5 m/sec
Kinetic energy become

Therefore energy lost due to friction = 72 -50 = 22 J
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