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3 years ago

A 0.580-kg rock is tied to the end of a string and is swung in a circle with a radius of 0.500 meters. The velocity of the rock

is 4.50 m/s. What is the centripetal force acting on the rock

Physics

1 answer:

gtnhenbr [62]3 years ago

5 0

$F = m \frac{v^2}{r}$

$F= 0.58 \frac{4.5^2}{0.5} = 23.49 N$

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sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)