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4 years ago

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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?

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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The groups are ;

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Explanation:

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A) $+1.67 rad/s^2$

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Substituting into the equation, we find the angular acceleration:

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And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

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And solving for $\Delta t$ we find

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$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

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Solving for $\theta$ ,

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Explanation:
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Volume, " V = ? ; (unknown, we need to solve for this).
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mass , "m" = 27.63 g ;
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The formula for density is:
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D = m / V ; Divide each side of the equation by: "(1/m)" ;
to isolate " V" on one side of the equation; and to solve for "V" ;
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1/m)*D = (1/m) * (m / V) ;
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to get:
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D/m = ; ↔ 1 / V ; Take the reciprocal of EACH SIDE; to isolate "V" on each side of the equation:
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m / D = V/1 ↔ V = m / D ;
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V = m / D ;
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The volume is: " 1.430 cm³ " .
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