Question

A pendulum is made up of a small sphere of mass 0.500 kgkg attached to a string of length 0.600 mm. The sphere is swinging back and forth between point AA, where the string is at the maximum angle of 35.0∘∘ to the left of vertical, and point CC, where the string is at the maximum angle of 35.0∘∘ to the right of vertical. The string is vertical when the sphere is at point BB. Calculate how much work the force of gravity does on the sphere from AA to BB.

Answer 1

Answer:

The work done by the force of gravity on the sphere from AA to BB is W=0.53 J.

Explanation:

The length of the string is 0.6 m.

We can define the work done by the force of gravity as:

$W=\int \bar{F} dy \approx F\Delta y=Fh$

The integral above is a integral of a vector product between the force of gravity and the displacement of the pendulum (dy).

In this case, we can simplify that because the force of gravity acts only in the y-axis, so the only displacement that afects our work is the displacement done in that direction (h).

We can calculate h as

$h=L*(1-cos(\alpha))=0.6(1-cos(35)=0.6(1-0.82)=0.6*0.18\\\\h=0.108\, m$

The force of gravity is

$F=mg=0.5\,kg*9.81\,m/s^2=4.905\,N$

Then the work done by the force of gravity is:

$W=F*h=4.905N*0.108m=0.53\,Nm=0.53\,J$

Answer 2

The number of work the force of gravity does on the sphere from AA to BB is W=0.53 J.

Calculation of the work done:

Since

A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.600 mm

So, here the direction is

= l*(1- cos)

= 0.6 * (1 - cos(35)

= 0.6 * (1 - 0.82)

= 0.6*0.18

= 0.108m

Now the force of gravity should be

= 0.5* 9.81

= 4.905 N

Now the work done should be

= f * h

= 4.905 N * 0.108 m

= 0.53 Nm

= 0.53J

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