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A 20 gram bullet exits the rifle at a speed of 985 m/s. the gun barrel is 0.8 m long. what
was the force during the rocket's launch?


Sagot :

leena

Hello!

We can use the following relationship:

[tex]\Delta KE = W = F \cdot d[/tex]

ΔKE = Change in Kinetic Energy (J)
W = Work done on object (J)

F = Force (N)

d = distance of which the force is applied to the object. (This is equivalent to the length of the gun barrel, or 0.8 m)

Also, recall that:
[tex]KE = \frac{1}{2}mv^2[/tex]

m = mass of bullet (0.02 kg)

v = velocity of bullet (vfinal = 985 m/s, vinitial = 0 m/s)

We can now solve.

[tex]\Delta KE = F \cdot d\\\\KE_f - KE_i = F \cdot d\\\\\frac{1}{2}(0.02)(985^2) - \frac{1}{2}(0.02)(0^2) = (0.8)F\\\\9702.25 = 0.8F\\\\F = \frac{9702.25}{0.8} = \boxed{12127.813 N}[/tex]

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