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A single constant force F= (3i+5j) N acts ona 4.00-kg particle.

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asked 2020-10-23

A single constant force \(F= (3i+5j) \) N acts ona 4.00-kg particle. (a) Calculate the work done by this forceif the particle moves from the origin to the point having the vector position \(r = (2i-3j) m\). Does thisresult depend on the path? Explain. (b) What is the speed ofthe particle at r if its speed at the origin is 4.00 m/s? (c) What is the change in the potential energy?

Answers (1)

2020-10-24

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

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