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saveliy_v [14]
4 years ago
8

A block rests on a horizontal frictionless surface. A string is attached to the block, and is pulled with a force of 46.0 N at a

n angle θ above the horizontal. After the block is pulled through a distance of 2.00 m, its speed is 3.10 m/s, and 42.0 J of work has been done on it. What is the angle θ?

Physics
1 answer:
lord [1]4 years ago
7 0

Answer:

The angle is 62.83 degrees

Explanation:

Hello!

Briefly, the concept of work is defined as the energy needed to move a certain distance body, it is calculated as the product of the force in the direction of movement by the distance traveled.

Taking into account the above, this exercise is resolved with the following steps.

1. draw a problem diagram involving the forces and angle (see attached image)

2. As you can see only the horizontal component (fcos) is doing work, then the equation for the work of this body is given by the horizontal force by the distance traveled as follows

W=FcosΘd

Where

W=work=42J

F=force=46N

d=distance traveled=2m

now we use algebra to find the angle, and use the values of force, distance, and work

\frac{W}{Fd} =cos\alpha

\alpha =cos^-1(\frac{W}{Fd} )=cos^-1(\frac{42}{(46)(2)} )=62.83

The angle is 62.83 degrees

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2. An 873 kg dragster accelerates at a rate of 44.6 m/s during a race.
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8 0
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A man stands on a cliff that 10m above the sea. He throws a stone vertically upwards with velocity 5ms^-1. The stone eventually
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Position =
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H = starting position + (starting speed x t) + 1/2 A t²

Here's how we can use it, with some careful definitions:

-- Let's say the surface of the sea is zero height.
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--  Starting speed is +5 ... 5 m/s upward, when he tosses it.

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I think this is going to work out just beautifully !

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Solve it for 't' with the quadratic equation,
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When you solve a quadratic with the formula, you always get two roots.
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The two roots are

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Answer:

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Hence the acceleration is 0.1m/s²

3 0
3 years ago
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