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3 years ago

10

A horizontal force of 12 Newtons is applied to a 4.0 kg box that slides on a horizontal surface. The box starts from rest moves

a horizontal distance of 10 meters and obtains a velocity of 5.0 m/s. The surface has friction. The friction force is
Group of answer choices

4.9 N.

6.0 N.

7.0 N.

6.5 N.

1 answer:

Serjik [45]3 years ago

4 0

Answer:

7.0N

Explanation:

F=ma

v2=u2+2as

25=0+20a

a=1.25m/s2

F=ma

F=0.4×1.25

Fn=5N

F=12

12-5=Ff

=7.0N

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the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during

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Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

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$9v_1 = 363 - 2v_2$

$v_1=\frac{363 - 2v_2}{9}$

The mathematical expression for the conservation of kinetic energy is

$\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2$

$\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2$ ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$

$21681 = 9v_1^2+2v_2^2$

Substituting (i) in (ii)

$21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2$

$(363-2v_2)^2+18v_2^2=195129$

$(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0$

$22v_2^2-145v_2-63360=0$

Solving the equation, we get

$v_2=96 \ m/s, -30 \ m/s$

The negative velocity is neglected.

Therefore, substituting 96 m/s for $v_2$ in (i), we get

$v_1=\frac{363-(2 \times 96)}{9}$

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Answer:

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Answer:

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