An object with a mass of #200 kg# is hanging from an axle with a radius of #6 cm#. If the wheel attached to the axle has a radius of #15 cm#, how much work would it take to turn the wheel a length equal to the circumference of the axle?

Question

An object with a mass of #200  kg# is hanging from an axle with a radius of #6 cm#.  If the wheel attached to the axle has a radius of #15 cm#, how much work would it take to turn the wheel a length equal to the circumference of the axle?

Julia Aguilar · Accepted Answer

The work done is #=295.6J#

The load L=(200g)N#

Radius of axle #r=0.06m#

Radius of wheel #R=0.15m#

Taking moments about the center of the axle

#F*R=L*r#

#F*0.15=200*g*0.06#

#F=(200*g*0.06)/0.15=784N#

The circumference of axle is #d=2pir=0.12pi#

The work done is

#W=Fd=784*0.12pi=295.6J#

Nicholas Agrusa · Answer

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

$$ W = F \times d $$

Where:
- $ W $ is the work done
- $ F $ is the force applied
- $ d $ is the distance over which the force is applied

First, we need to calculate the force applied to turn the wheel. This can be found using the torque formula:

$$ \tau = r \times F $$

Where:
- $ \tau $ is the torque
- $ r $ is the radius
- $ F $ is the force applied

Given that the wheel attached to the axle has a radius of 15 cm, and the axle itself has a radius of 6 cm, we can calculate the torque applied:

$$ \tau = (15 \, cm + 6 \, cm) \times F $$

$$ \tau = 21 \, cm \times F $$

Next, we can calculate the force applied:

$$ F = \frac{\tau}{r} $$

$$ F = \frac{21 \, cm \times F}{6 \, cm} $$

$$ F = \frac{21}{6} \, F $$

$$ F = 3.5 \, F $$

Now, we can calculate the work done:

$$ W = F \times d $$

Given that the distance $ d $ is equal to the circumference of the axle, which is $ 2\pi \times 6 \, cm $, we have:

$$ W = 3.5 \, F \times 2\pi \times 6 \, cm $$

$$ W = 21\pi \, F \, cm $$

So, the work required to turn the wheel a length equal to the circumference of the axle is $ 21\pi \, F $ joules.