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4 years ago

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A 5.00-kg box slides 4.00 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floo

r and the box if the box had an initial speed of 3.00 m/s?

1 answer:

zloy xaker [14]4 years ago

6 0

Explanation:

According to the given situation,

work done = change in kinetic energy

As the formula of kinetic energy is as follows.

K.E = $\frac{1}{2}mv^{2}$

Putting the given values into the above formula as follows.

K.E = $\frac{1}{2}mv^{2}$

= $\frac{1}{2} \times 5 kg \times (3 m/s)^{2}$

= 22.5 J

Also we know that,

Work done = force x distance

or, force = $\frac{work}{distance}$

= $\frac{22.5 J}{4 m}$

= 5.62 N

In the given case, force is friction and formula to calculate friction is as follows.

friction = $\mu \times m \times g$

where, $\mu$ = the coefficient of friction

Putting the given values into the above formula as follows.

Friction = $\mu \times m \times g$

5.62N = $\mu \times 5 \times 9.8$

$\mu$ = 0.114

Thus, we can conclude that the coefficient of kinetic friction between the floor and the given box is 0.114.

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The electric field created by Q at the position of q is $\frac{F}{Q}$ .

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The electric field created by Q at the position of q is calculated as follows;

$E = \frac{F}{Q} \\\\$

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If an object has an increasing positive acceleration, what can we assume about the forces on it?

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What is the the acceleration of a proton that is 4.0 cm from the center of the bead? input positive value if the acceleration is

QveST [7]

Missing detail in the text:
"<span>A small glass bead has been charged to + 25 nC "

Solution
The force exerted on a charge q by an electric field E is given by
</span> $F=qE$
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</span> $E=k_e \frac{Q}{r^2}$
with $k_e = 8.99\cdot 10^9 Nm^2/C^2$ , $Q=+25 nC=25 \cdot 10^{-9}C$ is the charge on the bead. We want to calculate the field at $r=4.0 cm=0.04 m$ :
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The proton has a charge of $q=1.6\cdot 10^{-19}C$ , therefore the force exerted on it is
$F=qE=1.6\cdot 10^{-19}C \cdot 1.4\cdot 10^5 V/m=2.25\cdot 10^{-14} N$

And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass, $m=1.67\cdot 10^{-27} kg$ , we have
$F=ma$
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4 years ago

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disa [49]

Answer:

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Explanation:

a) The magnitude of the change in the ball's momentum is:

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$\Delta p = 1.350\,\frac{kg\cdot m}{s}$

b) The change in the magnitude of the ball's momentum:

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$\Delta p' = -0.454\,\frac{kg\cdot m}{s}$

c) The magnitude of the change in the ball's momentum is more directly related to the net force acting on the ball, as it measures the effect of the force on change in ball's motion at measured time according to the Impact Theorem. So, the right answer is option D.

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