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3 years ago

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A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h

ave negligible friction, and the rope does not slip on the pulley. On one end of the rope hangs a 3.00kg bunch of bananas, and on the other end hangs a 4.50kg monkey. Calculate the downward acceleration of the monkey and the tension in both ends of the rope.

1 answer:

boyakko [2]3 years ago

4 0

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α

where, α is the angular acceleration

α = a / R

(T2 - T1)R = 0.5 x m3 x R² x a / R

T2 - T1 = 0.5 x m3 x a ..... (3)

from (1), (2) and (3)

$a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g$

$a = \frac{4.5-3}{3+4.5+0.75}\times 9.8$

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

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A plane coming in to land at a busy airport is asked to circle the airport until the air traffic congestion eases off. The pilot

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Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

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VMariaS [17]

Answer:

F=ma

here F is force, m is mass and a is accelaration,

According to the question,

F=3*F= 3F

m= 1/3 of m= m/3

a= ?

so the equation becomes,

3F= m/3*a

3F*3= ma

9F=ma

F= ma/9

Therefore accelaration reduces by 1/9.

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