Question 6.1: WORK DONE BY A CONSTANT FORCE (a) Steve exerts a steady forc...

IDENTIFY and SET UP:

In both parts (a) and (b), the target variable is the work W done by Steve. In each case the force is constant and the displacement is along a straight line, so we can use Eq. (6.2) (W=F s \cos \phi) or (6.3) (W=\vec{F} \cdot \vec{s}). The angle between \overrightarrow{\boldsymbol{F}} and \overrightarrow{\boldsymbol{s}} is given in part (a), so we can apply Eq. (6.2) (W=F s \cos \phi) directly. In part (b) both \overrightarrow{\boldsymbol{F}} and \overrightarrow{\boldsymbol{s}} are given in terms of components, so it’s best to calculate the scalar product by using Eq. (1.19): \overrightarrow{\boldsymbol{A}} \cdot \overrightarrow{\boldsymbol{B}}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}.

EXECUTE:

(a) From Eq. (6.2): W=F s \cos \phi=(210 \mathrm{~N})(18 \mathrm{~m}) \cos 30^{\circ}=3.3 \times 10^{3} \mathrm{~J}

(b) The components of \overrightarrow{\boldsymbol{F}} \text { are } F_{x}=160 \mathrm{~N} \text { and } F_{y}=-40 \mathrm{~N}, and the components of \overrightarrow{\boldsymbol{s}} are x = 14 m and y = 11 m. (There are
no z-components for either vector.) Hence, using Eqs. (1.19) and (6.3), we have

W=\overrightarrow{\boldsymbol{F}} \cdot \overrightarrow{\boldsymbol{s}}=F_{x} x+F_{y} y= (160 N)(14 m) + (-40 N)(11 m) =1.8 \times 10^{3} \mathrm{~J}

EVALUATE: In each case the work that Steve does is more than 1000 J. This shows that 1 joule is a rather small amount of work.