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DiKsa [7]
3 years ago
7

to masses 7 kg and 12 KG are connected at the two ends of a light inextensible string that passes over a fictional Pulley using

free body diagram method find acceleration of masses and the tension in the string when the masses are released ​
Physics
1 answer:
Alex17521 [72]3 years ago
8 0

Tension in the string when the masses are released ​is 88.42  N

Acceleration of masses is \bold{a=2.578 m/sec^2}

Explanation:  

Given:

Mass ,m1 = 12

Mass , m2 = 7  

g = 9.8m/s^2

To Find :

Tension  in the string=?

Acceleration of masses=?

Solution:

For mass M_1  

M_1 a=T-M_1 g--------------------(1)

For mass M2

M_2 a=T-M_2 g---------------------(2)

Adding equation (1) and (2)

(M_1+M_2)a=(M_2-M_1)g

Finding Acceleration:

Acceleration is given by

a=(M_2-M_1 )/(M_1+M_2) g

Substituting the values,

a=\frac{(12-7)(9.8)}{(7+12)}

a=\frac{(5)(9.8)}{19}

a=\frac {49}{19}

a=2.578 m/sec^2

Finding Tension:

From  Equation 1

M_1 a=T-M_1 g

Tension can be

T=M_1 a+M_1 g

T=(7)(2.578) + 7(9.8)

T=(17.99)+(68.6)

T=86.59 N

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Answer:

6.51 m and 37.13 m

Explanation:

from the question we were given

mass of skeet = 0.25 kg

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angle = 30 degrees to the horizon

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speed of pellet = 2000 m/s

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Vx = 25 cos 30 = 21.65

Vy = 25 sin 30 = 12.5

now

V = U + (a x t)

where V = final velocity, U = initial velocity , a = acceleration, t = time and s = distance

-25 sin 30 = 25 sin 30 + (-9.8 x t)

-12.5 = 12.5 - 9.8 t

t = 2.55 s

also

V^2 = U^2 + 2as  ( s = vertical distance and V = 0 )

0 = (25 sin 30)^2 + 2 x (-9.8) x Y

19.6 Y = 156.25

Y =7.97 m

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distance = 21.65 x 2.55 = 55.2 m

applying the conservation of linear momentum

on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx)  ...equation 1

on the y axis :   (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2

(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx

 Vx = 20.42m/s

0 + (0.015 x 200) = (0.25 + 0.015) Vy

 Vy = 11.32 m/s

now V^2 = U^2 + 2 as

 0 = 11.3^2 + (2 x (-9.8) x s)

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  • to find the extra distance moved after collision we apply

s = ut + 1/2 at^2

-7.98 = 11.32t + 1/2 (-9.8) t^2

4.9 t^2 - 11.32t + 7.98

t =  3.17 s

recall that distance = speed x time

distance = 20.42 x 3.17 = 64.73 m

the distance of the skeet before being hit would be half of the distance it travels without being hit, this is because the skeet was hit at its maximum height = 55.2 /2

= 27.6 m

therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m

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3 years ago
A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti
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Answer:

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Explanation:

Given;

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final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

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