A 3.0 g bullet traveling at a speed of 400 m/s enters a tree and exits the other side with a speed of 200 m/s. Where did the bullet's lost KE go, and what was the energy transferred?

1800J

Explanation:

Step one:

given data

mass of bullet m= 3g= 0.03kg

initial velocity u = 400m/s

final velocity v= 200 m/s

Step two:

1.The bullet's lost kinetic energy went inside the tree.

2. The energy transferred is computed as

= initial KE- KE final

Initial KE= 1/2mu^2

Initial KE= 1/2*0.03*400^2

Initial KE= 1/2*0.03*160000

Initial KE= 1/2*4800

Initial KE= 2400J

KE final= 1/2mv^2

KE final= 1/2*0.03*200^2

KE final= 1/2*0.03*40000

KE final= 1/2*1200

KE final= 600J

KE transferred = 2400-600

KE transferred= 1800J

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