How much kinetic energy does a 0.30-kg stone have if it is thrown at 44 m/s?
2 answers:
You might be interested in
Answer:
t=17.838s
Explanation:
The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:
The acceleration is, by definition:
So, the velocity can be obtained by integrating this expression:
The velocity is, by definition: , so
.
Do x=11 in order to find the time spent.
At this time the velocity is:
This velocity remains constant in the section 2, so for that section the movement equation is:
The left distance is 89 meters, and the velocity is , so:
So, the total time is 14.303+3.5355s=17.838s
Answer:
The answer to the question is
The roller coaster will reach point B with a speed of 14.72 m/s
Explanation:
Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h
Where m = mass
g = acceleration due to gravity = 9.81 m/s²
h = starting height of the roller coaster
we have the given variables
h₁ = 36 m,
h₂ = 13 m,
h₃ = 30 m
v₁ = 1.00 m/s
Total energy at point 1 = 0.5·m·v₁² + m·g·h₁
= 0.5 m×1² + m×9.81×36
=353.66·m
Total energy at point 2 = 0.5·m·v₂² + m·g·h₂
= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m
The total energy at 1 and 2 are not equal due to the frictional force which must be considered
Total energy at point 2 = Total energy at point 1 + work done against friction
Friction work = F×d×cosθ = ( × mg)×60×cos 180 = -117.72m
0.5·m·v₂² + 127.53·m = 353.66·m -117.72m
0.5·m·v₂² = 108.41×m
v₂² = 216.82
v₂ = 14.72 m/s
The roller coaster will reach point B with a speed of 14.72 m/s
