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3 years ago

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A boy takes hold of a rope to pull a wagon (m = 50 kg) on a surface with a static coefficient of friction μS = 0.25. Calculate t

he force (in newtons) that would need to be applied to the rope to just start the wagon moving.

2 answers:

vivado [14]3 years ago

6 0

Answer:

<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>

Explanation:

Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.

F = μ x mg .............................. 1

where F is the frictional force;

μ is the coefficient of friction ( μs, in this case, static friction);

m is mass of the object and;

g is the acceleration due to gravity( a constant equal to 9.81 m/ $s^{2}$ )

from the equation we are provide with;

μs = 0.25

m = 50 kg

g = 9.81 m/ $s^{2}$

F =?

Using equation 1

F = 0.25 x 50 kg x 9.81 m/ $s^{2}$

F = 122.63 N

<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>

brilliants [131]3 years ago

5 0

Explanation:

Below is an attachment containing the solution.

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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

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Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

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mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

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Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

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$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

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Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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