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4 years ago

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In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

ntal frictionless bearings, has a radius of 1.70 cm. When released from rest, the heavier block is observed to fall 60.6 cm in 7.00 s (without the string slipping on the pulley).

1 answer:

Wittaler [7]4 years ago

5 0

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

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

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

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a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

$x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}$ (1)

Where:

$x_{o}$ - Initial x-position, in meters.

$v_{o,x}$ - Initial x-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{x}$ - x-acceleration, in meters per second.

If we know that $x_{o} = 0\,m$ , $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}$

$x(t) = 0.324\,m$

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

$y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}$ (2)

Where:

$y_{o}$ - Initial y-position, in meters.

$v_{o,y}$ - Initial y-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{y}$ - y-acceleration, in meters per second.

If we know that $y_{o} = 0\,m$ , $v_{o,y} = -3.6\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{y} = 0\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}$

$y(t) = -2.16\,m$

The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboarder ( $v_{x}$ ), in meters per second, is calculated by this kinematic formula:

$v_{x}(t) = v_{o,x} + a_{x}\cdot t$ (3)

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-velocity of the skateboarder is:

$v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)$

$v_{x}(t) = 1.08\,\frac{m}{s}$

The x-velocity of the skateboard is 1.08 meters per second.

d) As the skateboarder has a constant y-velocity, then we have the following answer:

$v_{y} = -3.6\,\frac{m}{s}$

The y-velocity of the skateboard is -3.6 meters per second.

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