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2 years ago

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A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo

rce diagram and determine the net force. Then calculate the acceleration of the object.

1 answer:

Taya2010 [7]2 years ago

3 0

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be $a$ m/s².

Given:

Mass of the block is, $m=5\ kg$

Force of pull is, $F=40\ N$

Frictional force on the block is, $f=8\ N$

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

$F_{net}=F-f=40-8=32\ N$

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

$F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2$

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

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Answer:

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2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

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$K=U$

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With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

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Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

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part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

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So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

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Answer:

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Explanation:

given data

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