Question

A 7.80-g bullet moving at 540 m/s strikes the hand of a superhero, causing the hand to move 5.10 cm in the direction of the bullet's velocity before stopping. (a) Use work and energy considerations to find the average force that stops the bullet.

Answer 1

Answer:

$F = 22298.824\,N$

Explanation:

According to the Principle of Energy Conservation and the Work-Energy Theorem, the bullet has the following expression:

$U_{g,A} + K_{A} = U_{g,B} + K_{B} + W_{loss}$

$W_{loss} = U_{g,A}-U_{g,B} + K_{A}-K_{B}$

$F\cdot \Delta s = \frac{1}{2}\cdot m \cdot [v_{A}^{2}-v_{B}^{2}]$

The average force exerted on the bullet to stop it is:

$F = \frac{m\cdot [v_{A}^{2}-v_{B}^{2}]}{2\cdot \Delta s}$

$F = \frac{(7.8\times 10^{-3}\,kg)\cdot [(540\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}]}{2\cdot (0.051\,m)}$

$F = 22298.824\,N$

Answer 2

Answer:

22298.82N

Explanation:

The bullet has a kinetic energy: ½*m*v²

mass of bullet = 7.80g = 7.80÷1000 = 0.0078kg

distance in meter =5.10cm = 5.10÷100 = 0.051meter

K.e = 0.5 ×0.0078 × 540× 540

K.e = 1137.24 J

When the bullet stops this energy goes to zero, as the energy must be conserved the work done by the head of the superhero must be equal to the original energy of the bullet:

W= K

Considering an average constant force, the work can be calculated as:

W=F*d=K

Solving for F:

F = K/d

1137.24 J/ 0.051m

= 22298.82N