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Nata [24]
3 years ago
7

A 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a)During one complete circle, starti

ng anywhere, calculate the total work done on the ballby (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along thesemicircle from the lowest to the highest point on the path.
Physics
1 answer:
marishachu [46]3 years ago
6 0

Answer: a) i) 0 ii) 0  b) i) 0 ii) -25.1 J

Explanation:

a) For the first part, as work is defined as the process through an applied force produces a displacement in the direction of the movement, if the start and the end points are the same (as for a complete circle), this means that there is no displacement, i.e. total work (for any applied force) is 0.

b) Taking only a semicircle from the lowest to the highest point on the path, as tension force is always perpendicular to the displacement, we can conclude that the tension force does no work.

Now, for gravity, the displacement, is just the difference in height between the highest and lowest point, which is equal to twice the length of the string, and is a vector with only a vertical component, as follows:

Δy = y₂ - y₁ (by definition of displacement)

Now, the work done by gravity is just the product of the weight of  the ball, times the vertical displacement (as the work done by gravity is independent of the trajectory) , as follows :

W = m. g. (y₂-y₁) = 0.8 Kg. 9.8 m/s². (-2. 1.6 m) = -25.1 J

(The minus sign is due to gravity and the displacement point in opposite directions, in this case we take as positive the downward direction).

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2 years ago
A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
  • Mass of the box, m = 3.1 kg
  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
  • Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

Learn more about work done here: brainly.com/question/8119756

8 0
2 years ago
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