energy is zero at the initial position. So we have 0=-mg(43.6+x)+k →x2-2.255x-9832=0→x=11.1 4. The work done by the gravitational force in moving a system of mass m at a distance r from the center of a uniform sphere(of mass GMm M) to a distance reis ) where both ri and rg R e less than the radius R of the sphere(see Figure 4). This imply about the work done by this form of the gravitational force around a closed path is zero Solution Since the mass is in the uniform sphere, te gravitational force is a function of position r, so the gravitational force is GmM 4 FG △ →W=dF=( GMm r)·drr=- R 5. a particle moves along the x axis under the influence of a conservative force that is described by F=-axe where a and are constants. The potential energy function U(x)is 2B Solution According to the definition of the potential energy, we ge U(x)=[.d=-areeaxdxsareaxar 3peaai0eva2 6. A force F=-kxi-hy is one of the forces on a particle of mass m. The quantity k is constant This force represents a two-dimensional Hooke's force law. Is this force conservative?yes The appropriate potential energy function for this force is -k(x+y)energy is zero at the initial position. So we have 2.255 98.32 0 11.1 2 1 0 (43.6 ) 2 2 ⇒ − − = ⇒ = = − + + x x x mg x kx 4. The work done by the gravitational force in moving a system of mass m at a distance ri from the center of a uniform sphere (of mass M) to a distance rf is ) 2 2 ( 2 2 3 i rf r R GMm − − , where both ri and rf are less than the radius R of the sphere (see Figure 4). This imply about the work done by this form of the gravitational force around a closed path is zero . Solution: Since the mass is in the uniform sphere, te gravitational force is a function of position r, so the gravitational force is rr R GMm r r R M r Gm FG ˆ ˆ 3 4 3 4 3 3 3 2 = − ⋅ ⋅ π = − ⋅ π r ) 2 2 ( 2 d ( )ˆ dr 2 2 3 2 3 3 | i f r r r r f i G r r R r GMm R GMm r r r R GMm W F r f i f i ⇒ = ⋅ = − ⋅ ⋅ = − ⋅ = − − ∫ ∫ r r r 5. A particle moves along the x axis under the influence of a conservative force that is described by F xe i x ˆ 2 β α − = − r , where α and β are constants. The potential energy function U(x) is 2 2 x e β β α − . Solution: According to the definition of the potential energy, we get 2 2 2 2 2 2 d 2 ( ) d d 2 x x x x x x x p U x F r xe x e x e e β β β β β α β α α α − ∞ − ∞ − ∞ − ∞ = ⋅ = − = − = − = ∫ ∫ ∫ v r 6. A force F kxi kyj = − ˆ − ˆ r is one of the forces on a particle of mass m. The quantity k is constant. This force represents a two-dimensional Hooke’s force law. Is this force conservative? yes . The appropriate potential energy function for this force is ( ) 2 1 2 2 k x + y . m R Fig.4 ir r f r r