Force is given as \(F=(3i+5j)N\ mass= 4kg\)

a) workdone \(= F.ds= (3i+5j).(2i-3j)=6-15=-9 J\)

the work done is path independant since we did not considerany frictional force.

b) speed at r if speed at origin is 4 m/s

we can calculate acceleration \(\displaystyle{a}={\frac{{{\left|{F}\right|}}}{{{m}}}}=?\frac{{{32}+{52}}}{{4}}={1.46}\frac{{m}}{{s}^{{2}}}\)

we can use \(\displaystyle{v}^{{2}}={u}^{{2}}+{2}{a}{s}\) with s beings \(\displaystyle=?{\left({22}+{32}\right)}={3.6}{m}\)

\(\displaystyle{v}^{{2}}={4}^{{2}}+{2}{\left({1.46}\right)}{\left({3.6}\right)}\)

solve for v

c)You can use energy conservation to get this part