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What average force is needed to accelerate a 9.20-gram pellet from rest to 125 m/s over a distance of 0.800 m along the barrel of a rifle?

Respuesta :

Given:
u = 0, initial velocity
v = 125 m/s, final velocity
s = 0.0800 m, distance traveled
9.20 g, the mass of the pellet

If the acceleration is a, then
0² + 2*(a m/s²)*(0.800 m) = (125 m/s)²
1.6a = 15625
a = 9765.625 m/s²

Calculate the force.
F = (9.20 x 10⁻³ kg)*(9765.625 m/s²) = 98.84 N

Answer: 98.8 N (nearest tenth)

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