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A rifle bullet with a mass of 17.5 g traveling toward the right at 293 m/s strikes a large bag of sand and penetrates it to a depth of 23.6 cm. Determine the magnitude and direction of the friction force (assumed constant) that acts on the bullet.

3 answers

a = (V^2-Vo^2)/2d.
a=(0-(293)^2)/0.472m = -5.5*10^6 m/s^2.

Ff = a/g = -5.5*10^6/9.8 = -5.6*10^5 N.
To the left.
Correction:
a/g = Coefficient of friction,u.
u = a/g = = -5.5*10^6/9.8 = -5.6*10^5.

Ff = u*m*g = -5.6*10^5*0.0175*9.8 = -9.6*10^4 N. To the Left.
Correction:

a = (V^2-Vo^2)/2d.
a = (0-(293)^2)/0.472m = -181,883 m/s^2.
u = a/g = -181883/9.8 = -18,560.

Ff = u*mg = 18560*0.0175*9.8 = 3183 N. To the left.
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