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The three spheres in Fig. 13-45, with massesmA=80kg, mB=10g, and mC=20g , have their centers on a common line,withL=12cm and d=4.0cm . You move sphereBalong the lineuntil its center-to-center separation fromCis d=4.0cm . Howmuch work is done on sphereB(a) by you and (b) by the net gravitational force onBdue to spheresAandC?

Short Answer

Expert verified
  1. The work done on sphere B by you is +5.0×10-13J.
  2. The work done on the sphere by the net gravitational force on due to Sphere A and C is -5.0×10-13J.

Step by step solution

01

Step 1: Given

mA=80g,mB=10g,mC=20gL=12cmd=4.0cm

02

Determining the concept

Using the formula for potential energy, find the potential energy of the system. By using the formula for work done, find thework done on sphere B by us and by the net gravitational force on due to Sphere Aand C.

Formulae are as follows:

U=-GmAmBrW=Uf-Ui

where mA,  mBare masses, r is the radius, G is gravitational constant, W is work done and U is potential energy.

03

(a) Determining the work done on sphere b by us

Now,

U=-GmAmBr

Therefore, initial potential energy is,

Ui=-GmAmBd-GmAmCL-GmBmCL-d

The final potential energy is,

Uf=GmAmBL-d-GmAmCL-GmBmCd

Hence, work done can be found as,

W=Uf-UiW=GmAmBL-d-GmAmCL-GmBmCd--GmAmBd-GmAmCL-GmBmCL-dW=GmBmA1d-1L-d+mC1L-d-1dW=GmBmAL-2ddL-d+mC2d-LdL-dW=GmBmA-mCL-2ddL-dW=6.67×10- 11m3s2·kg0.010Kg0.080kg-0.020kg0.12m-20.040m0.040m0.12-0.040mW=+5.0×10-13J

Hence,the work done on sphere B by you is +5.0×10-13J.

04

(b) Determining the work done on the sphere by the net gravitational force on   due to the sphere  A and C

Now,

W=Uf-Ui

Therefore, work done by gravity is,

W=-Uf-UiW=-5.0×10-13J

Hence, the work done on the sphere by the net gravitational force on B due to Sphere A and C is-5.0×10-13J .

Therefore, using the formula for potential energy, the work done can be found.

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