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3 years ago

A student pulls on a 20 kg box with a force of 50 N at an angle 45 degrees relative to the horizontal. The box increases

Physics

1 answer:

telo118 [61]3 years ago

8 0

The value of the friction force is 5.4 N

Explanation:

In order to solve the problem, we have to write down the equation of the forces along the horizontal direction.

We have only two forces acting along the horizontal direction:

The horizontal component of the pull of the student, $F cos \theta$ , forward
The friction force, $F_f$ , backward

The resultant force must be equal to the product between mass and acceleration (Newton's second law), so we can write:

$F cos \theta - F_f = ma$

where:

F = 50 N is the force of pull of the student

$\theta=45^{\circ}$ is the angle relative to the horizontal

m = 20 kg is the mass of the box

$a=1.5 m/s^2$ is the acceleration

Solving the equation for $F_f$ , we find the value of the friction force:

$F_f = Fcos \theta -ma=(50)(cos 45^{\circ})-(20)(1.5)=5.4 N$

Learn more about friction:

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Answer:

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Explanation:

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$g$ - Gravitational acceleration, measured in meters per square second.

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Please notice that $m\cdot g$ is the weight of the particle, measured in newtons.

a) If we know that $m = 3\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $z_{o} = 0\,m$ and $z_{f} = 4.1\,m$ , then the change in potential energy is:

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b) If we know that $m\cdot g = 650\,N$ , $z_{o} = 4.1\,m$ and $z_{f} = 0\,m$ , then the change in potential energy is:

$\Delta U_{g} = (650\,N)\cdot (0\,m-4.1\,m)$

$\Delta U_{g} = -2665\,J$

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