Question

A 300 g mass has a velocity of (3i+4j)m/sec at a certain instant .what is its kinetic energy

Answer 1

The Kinetic Energy of an object =1212 ⋅m⋅v2⋅m⋅v2

In this problem, velocity is 3i+4j3i+4j m/sm/s

Let's calculate the magnitude of this velocity. It will be =32+42−−−−−−√=5m/s=32+42=5m/s

Thus, kinetic energy in this case, will be

12⋅300g⋅52m2/s212⋅300g⋅52m2/s2

Conversion to the SI units gives;

12⋅0.3Kg⋅52m2/s212⋅0.3Kg⋅52m2/s2

=7.52⋅Kgm2/s2=7.52⋅Kgm2/s2

=3.75Joules=3.75Joules(Answer)

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Answer 2

velocity=√(3^2+4^2)=√25=5m/s
mass=.3kg
kinetic energy at that instant=.5*m*v^2
=.5*.3.25joules
=3.75 joules