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3 years ago

A 205 kg log is pulled up a ramp by means of a rope that

Physics

1 answer:

Volgvan3 years ago

7 0

Answer:2.737 kN

Explanation:

Given

mass of log(m)=205 kg

ramp inclination $=30^{\circ}$

coefficient of kinetic friction between log and ramp is ( $\mu _k$ )=0.9

log has an acceleration of 0.8 m/s^2

Let T be the tension in the rope

$T-mgsin\theta -f_r=ma$

Where $mgsin\theta$ =Sin component of weight

$f_r=friction\ Force=\mu _KN$ (Where N is Normal reaction)

$T-mgsin\theta -\mu _k\left ( mgcos\theta \right )=ma$

$T=m\left ( gsin\theta +\mu _kcos\theta +a\right )$

sin30=0.5

cos30=0.866

$T=205\times \left ( 9.81\times 0.5+0.9\times 9.81\times 0.866+0.8\right )$

$T=205\times 13.35=2.737 kN$

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2 years ago

A daring 510-N swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the t

oee [108]

Answer:

$v_x = 1.26 m/s$

Explanation:

given,

weight of swimmer = 510 N

length of ledge, L = 1.75 m

vertical height of the cliff, h = 9 m

speed of the swimmer = ?

horizontal velocity of the swimmer should be that much it can cross the wedge.

distance = speed x time

d = v_x × t

1.75 = v_x × t ........(1)

now,time taken by the swimmer to cover 9 m

initial vertical velocity of the swimmer is zero.

using equation of motion for time calculation

$s = ut +\dfrac{1}{2}gt^2$

$9= 0+\dfrac{1}{2}\times 9.8\times t^2$

t² = 1.938

t = 1.39 s

same time will be taken to cover horizontal distance.

now, from equation 1

1.75 = v_x × 1.39

$v_x = 1.26 m/s$

horizontal speed of the swimmer is equal to 1.26 m/s

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3 years ago

A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

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3 years ago