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4 years ago

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Starting from rest, a 88-kg firefighter slides down a fire pole. The average frictional force exerted on him by the pole has a m

agnitude of 800 N, and his speed at the bottom of the pole is 3.2 m/s. How far did he slide down the pole?

1 answer:

Cloud [144]4 years ago

3 0

Answer:h=7.22 m

Explanation:

Given

mass of firefighter $m=88 kg$

Frictional Force $F=800 N$

average speed at bottom $v=3.2 m/s$

Let h be the height of Pole

net force on Firefighter is

$F_{net}=88\times 9.8-800=62.4$

therefore net acceleration is

$a_{net}=\frac{62.4}{88}=0.709 m/s^2$

using $v^2-u^2=2ah$

here $v=3.2 m/s$

$u=0$

$(3.2)^2-0=2\times 0.709\times h$

$h=\frac{10.24}{1.418}$

$h=7.22 m$

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3 years ago

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Answer:

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b) $n_1=7.735 \,rev$

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Explanation:

Given that:

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time taken to reach full speed from rest, $t_1=11.9\,s$

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(a)

∵ 1 rev = 2π radians

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$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

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$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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4 years ago