An object with a mass of #8 kg# is hanging from an axle with a radius of #5 cm#. If the wheel attached to the axle has a radius of #25 cm#, how much work would it take to turn the wheel a length equal to the circumference of the axle?

Answer 1

The work is #=24.6J#

The load is #L=8gN#

The radius of the axle is #r=0.05m#

The radius of the wheel is #R=0.25m#

The effort is #=FN#

The acceleration due to gravity is #g=9.8ms^-2#

Taking moments about the center of the axle

#F*0.25=8g*0.05#

#F=8g*0.05/0.25=15.68N#

The force is #F=15.68N#

The distance is #d=2piR=2*pi*0.25=(0.5pi)m#

The work is #W=Fd=15.68*0.5pi=24.6J#

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Answer 2

To calculate the work required to turn the wheel a length equal to the circumference of the axle, we use the formula:

[ \text{Work} = \text{Force} \times \text{Distance} ]

The force exerted by the hanging object can be calculated using the formula for torque:

[ \text{Torque} = \text{Force} \times \text{Lever Arm} ]

where the lever arm is the radius of the axle. We can rearrange this formula to solve for force:

[ \text{Force} = \frac{\text{Torque}}{\text{Lever Arm}} ]

The torque exerted by the hanging object can be calculated as the product of its mass (8 kg) and the force of gravity (9.8 m/s^2), multiplied by the radius of the axle (0.05 m):

[ \text{Torque} = \text{mass} \times \text{gravity} \times \text{radius} ]

Substituting the given values:

[ \text{Torque} = 8 , \text{kg} \times 9.8 , \text{m/s}^2 \times 0.05 , \text{m} ]

[ \text{Torque} = 3.92 , \text{Nm} ]

Now, we can calculate the force exerted by the hanging object:

[ \text{Force} = \frac{3.92 , \text{Nm}}{0.05 , \text{m}} ]

[ \text{Force} = 78.4 , \text{N} ]

The distance the force acts over is the circumference of the axle, which is ( 2 \times \pi \times \text{radius of axle} ):

[ \text{Distance} = 2 \times \pi \times 0.05 , \text{m} ]

[ \text{Distance} = 0.314 , \text{m} ]

Now, we can calculate the work:

[ \text{Work} = \text{Force} \times \text{Distance} ]

[ \text{Work} = 78.4 , \text{N} \times 0.314 , \text{m} ]

[ \text{Work} = 24.57 , \text{J} ]

Therefore, it would take 24.57 Joules of work to turn the wheel a length equal to the circumference of the axle.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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