What is the kinetic energy in joules of a 0.02 kg bullet traveling 300 m/s?

Question

35922 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-18T12:21:06

#900 J#

Kinetic energy of bullet = #K.E=1/2 mv^2# where, #m# is the mass of the bullet and #v# is its velocity .

Given, #m=0.02Kg,v=300ms^-1#

So,#K.E =1/2 *(0.02)(300)^2=900J#

Answer 2 · 2018-03-18T12:32:23

#900 \ "J"#

Well, kinetic energy is expressed through the equation:

#"KE"=1/2mv^2#

#m# is the mass of the object in kilograms

#v# is the velocity of the object in meters per second

So, we plug in the values, which are

#m=0.02 \ "kg"#

#v=300 \ "m/s"#

And we get,

#"KE"=1/2*0.02 \ "kg"*(300 \ "m/s")^2#

#=1/2*0.02 \ "kg"*90000 \ "m"^2"/s"^2#

#=900 \ "kg m"^2"/s"^2#

#=900 \ "J"#