What is the scientific definition of energy that relates it to work
2 answers:
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<h2>Answer:</h2>
(a) 10N
<h2>Explanation:</h2>The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F -
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F - = m x a
F - = 50 x 0
F - = 0
F = -------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force is opposing the movement of the box.
∑F = 1.5F -
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F - = m x a
1.5F - = 50 x 0.1
1.5F - = 5 ---------------------(iii)
<em>Substitute </em><em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
<em />
The value of the force, F₀, at equilibrium is equal to the horizontal
component of the tension in string 2.
Response:
- The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>
Given:
The weight of the rod = The sum of the vertical forces in the strings
Therefore;
M·g = T₂·cos(37°) + T₁
The weight of the rod is at the middle.
Taking moment about point (2) gives;
M·g × L = T₁ × 2·L
Therefore;
Which gives;
F₀ = T₂·sin(37°)
Which gives;
- F₀ ≈ <u>0.377·M·g</u>
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Learn more about equilibrium of forces here: