Abullet with a mass of 3.0 gram, moving at speed of 300 m/s, hits and embeds itself in the trunk of a tree. the tree exerts an average force of 250 n on the bullet. what is the distance the bullet travels inside the trunk?

Explanation:

We have equation for force

          Force = Mass x Acceleration

          Force = 250 N

         Mass = 3 g = 3 x 10⁻³ kg  

         250 = 3 x 10⁻³ x Acceleration

         Acceleration = 83333.33 m/s²

Deceleration of bullet is 83333.33 m/s².

We have equation of motion v² = u² + 2as

Initial velocity, u = 300 m/s  

Acceleration, a = -83333.33 m/s²  

Final velocity, v = 0 m/s  

Substituting  

v² = u² + 2as

0² = 300² + 2 x -83333.33 x s

s = 0.54 m  

Distance the bullet travels inside the trunk is 0.54 m  

come closer and become far apart and become steam/gas

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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