Question

A bullet of mass 25g hits a wall with velocity 400m/s and emerges out with velocity 100m/s.Find work done.

Answer 1

Answer:

-1875 J

Explanation:

Given:

mass of bullet=m=25 gm

=0.025 kg

Initial velocity of bullet=u=400 m/s

Final velocity of bullet=v=100 m/s

Solution:

In this question we have to apply Work Energy Theorem which states that

'The total work done by all the forces is equal to the change in kinetic energy of the body'

and its formula is

W=ΔKE

  =KEf - KEi

In this case only the wall is doing work therefore the total work done by the system is equal to the work done by the wall only

KEi=1/2*m*u²

=1/2*0.025*(400)²

=2000 J

KEf=1/2*m*v²

=1/2*0.025*(100)²

=125 J

Plugging in the above values in the equation

W=ΔKE

=KEf - KEi

=125 - 2000

=-1875 J

Therefore the work done is -1875 J

(The negative sign indicates that the force and displacement are in opposite directions)

Hope this helps.

Answer 2

_________

Formulas Requied Are

_________

• 1 kg = 1000g
• $W= \frac{1}{2} m {v}^{2} - \frac{1}{2} m {u}^{2}$

Answer To Question ;-

As per given

m= 25g = 0.025 kg

u = 400 m/s

v = 100 m/s

According To Work Energy Theorem,

$work \: done \: = \: change \: in \: kinetic \: energhy \\ ie \\ </em><em>W</em><em> = \frac{1}{2} mv {}^{2} - \frac{1}{2} m {u}^{2} \\ </em><em>W</em><em> = \frac{1}{2} \times 0.025 \times {(100)}^{2} - \frac{1}{2} \times 0.025 \times {(400)}^{2} \\ </em><em>W</em><em> = \frac{1}{2} \times \frac{25}{1000} \times 10000 - \frac{1}{2} \times \frac{25}{1000} \times 160000 \\ </em><em>W</em><em> = 125 - 2000 \\ </em><em>W</em><em> = </em><em>-</em><em>1875 \: joule$