Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.
- The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>
Reasons:
The energy given to the block by the spring = 
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block =
= kinetic energy of
block = 
Therefore;
0.5·k·x² = 0.5·m·v²
Which gives;
x² ∝ v²
x ∝ v
On a plane inclined at an angle θ, we have;
The energy of the spring = 
- The force of the weight of the block on the string,

The energy given to the block =
= The kinetic energy of block as it leaves the spring = 
Which gives;

Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and <em>c</em> are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
- As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.
<em>Please find attached a drawing related to the question obtained from a similar question online</em>
<em>The possible question options are;</em>
- <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
- <em>The relationship is no longer linear and v will be more for the same value of x</em>
- <em>The relationship is still linear, with lesser value of v</em>
- <em>The relationship is still linear, with higher value of v</em>
- <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>
<em />
Learn more here:
brainly.com/question/9134528
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N