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2 years ago

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11. A box with a mass of 1.0 kg is resting on a horizontal surface and the coefficient of friction between the block and the sur

face is 0.20. It is accelerated by attaching a 1.5 kg mass, as shown in the diagram. Assume that the pulley is frictionless and that the cord has negligible mass.
a. What is the acceleration of the box? What is the tension?

2 answers:

Volgvan2 years ago

6 0

Answer:

A force is applied acting to the right. Assume that friction is negligible. Ikg 0.5N. For each question, one or more features of the system.

Explanation:hope this helps

AlladinOne [14]2 years ago

5 0

Answer:

We know that the force pulling the box in the positive x direction has a magnitude of m g sin 30 . Using Newtons Second Law, F = ma , we just need to solve for a :

ma=mgsin30

a=gsin30

=(10m/s2)(0.500)

=5m/s2

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Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

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time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

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t = time of observation

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b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

<u>Velocity just before opening the parachute:</u>

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

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$250=0+0.5\times 9.8\times t_h^2$

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<u>remaining time,</u>

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

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$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

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$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

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