What are the two reactants in this equation?
1 answer:
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The acceleration of the wagon is found by applying Newton's Second Law of motion.
1. The responses for question 1 are;
- x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
- y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
- x-component of the tension in the rope on the left is -80.0 N
- y-component of the tension in the rope on the left is 0
2. The net force in the x-direction is approximately <u>11.93 N</u>
3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.
Reasons:
The given parameters are;
Mass of the wagon, m = 100.0 kg
Angle of inclination to the horizontal of the rope to the right, θ = 40.0°
Tension in the rope on the right = 120.0 N
Direction in which the rope on the left is pulled = To the west
Tension in the rope on the left = 80.0 N
1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;
x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>
y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>
The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;
x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>
y-component = 80.0 N × sin(180°) = <u>0.0 N</u>
2. The net force in the horizontal direction, Fₓ, is found as follows;
Fₓ = The x-component of the rope on the left + The x-component of the rope on the right
Which gives;
Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>
3. The net acceleration of the block is given as follows;
According to Newton's Second Law of motion, we have;
Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ
Fₓ = m × aₓ
Therefore;
- The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.
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