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3 years ago

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Use this scenario to answer the four following questions. A person is on a sled on top of a snow covered hill. The vertical dist

ance from the top of the hill to the bottom is 12 meters. The sled and the person riding it have a combined mass of 60 kg. To start, person is pushed. The initial velocity is 1 m/s. At the bottom of the hill, the sled and the person have a velocity of 10 m/s. Use g=9.8 m/s2 or g=10 m/s2 for the gravitational acceleration. What is the initial total mechanical energy of the person and their sled?

1 answer:

jekas [21]3 years ago

7 0

Answer:7086 J

Explanation:

Given

height of hill from ground(h)=12 m

combined mass of sled and man(m)=60 kg

initial velocity(u)=1 m/s

at bottom velocity(v)=10 m/s

Total mechanical energy of the person

E=Potential Energy+Kinetic Energy

$E=mgh+\frac{mv^2}{2}$

$E=60\times 9.8\times 12+\frac{60\times 1^2}{2}$

E=7056+30=7086 J

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3 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

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

b) $n_1=7.735 \,rev$

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

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

e) $t_2=1.061\,s$

Explanation:

Given that:

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

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

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

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:

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

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

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

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

(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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3 years ago