#### To determine

**To find:**

The work done in moving the particle from *x=1* to *x=2* and interpret the answer by considering the work done from *x=1* to *x=1.5* and from *x=1.5* to *x=2*

#### Answer

_{0 J}

Interpretation: Work done from *x=1* to *x=1.5* is positive; it means kinetic energy of particle is increasing. Work done from *x=1.5* to *x=2* is negative; it means kinetic energy is decreasing.

#### Explanation

**1) Concept:**

Use the formula to calculate the work done.

**2) Formula:**

The work done in moving the object from *a* to *b* is given by

*W=∫abf(x)dx*

**3) Given:**

*fx=cos(πx3)*

**4) Calculation:**

The work done in moving the particle from *x=1* to *x=2* is given by using the formula

*W=∫12cos(πx3)dx*

Simplify,

*=sinπx3π312*

*=3πsinπx312*

*=3πsinπ(2)3-sinπ(1)3*

*=3π32-32*

*=0 N.m*

*=0 J*

Interpretation:

The work done in moving the particle from *x=1* to *x=1.5* is given by using the formula

*W=∫11.5cos(πx3)dx*

Simplify,

*=sinπx3π311.5*

*=3πsinπx311.5*

*=3πsinπ(1.5)3-sinπ(1)3*

*=3πsinπ2-sinπ(1)3*

*=3π1-32*

*=32π2-3 N·m*

*=32π2-3 J*

Work done is positive which means that kinetic energy is increasing.

Now,

The work done in moving the particle from *x=1.5* to *x=2* is given by using the formula

*W=∫1.52cosπx3dx*

Simplify,

*=sinπx3π31.52*

*=3πsinπx31.52*

*=3πsinπ(2)3-sinπ(1.5)3*

*=3πsin2π3-sinπ2*

*=3π32-1*

*=32π(3-2) N.m*

*=-32π2-3 J*

The force opposes the motion of the particle, the kinetic energy is decreasing, and hence, the work done is negative.

*W=∫12cosπx3dx=∫11.5cosπx3dx+∫1.52cosπx3dx*

*=32π2-3-32π2-3*

_{=0 J}

Therefore, the work done is *W=0 J*

**Conclusion:**

The work done is *W=0 J*

Interpretation: Work done from *x=1* to *x=1.5* is positive; it means kinetic energy is increasing. Work done from *x=1.5* to *x=2* is negative; it means kinetic energy is decreasing.