Question

A bullet of mass 0.010 kg and speed of 100 m/s is brought to rest in a wooden block

after penetrating a distance of 0.10 m. The work done on the bullet by the block is

A.

50 J.

B.

-

50 J.

C.

0.001 J.

D.

-

0.001 J.

E.

zero.

Answer 1

ANSWER:: -50 j or option (B)

A bullet of mass=m= 0.010 kg

A bullet of speed = v=100 m/s

d= distance of 0.10 m

W =work

F= force

KE = Kinetic energy

we know that,

   KE =1/2 mv² .............................(1)

and _{........................................2}

Where  =  is initial kinetic energy

And   ₌ final kinetic energy

Since the bullet brought to rest in a wooden block, the final kinetic energy of the bullet is 0. Substitute 0.010 kg for m and 100 m/s, we can say that  because the final position of bullet is rest so KE is zero

  

And ₌ (1/2)*m×v²

  =(1/2)*(0.010 kg)*(100 m/s)²

=50 J

W= 0 J- 50 J

=-50 J

the ans is = -50 J so (B) option is correct answer