Question

A 28 g bullet pierces a sand bag 29 cm thick. If the initial bullet velocity was 55 m/s and it emerged from the sandbag with 18 m/s. What is the magnitude of the friction force (assuming it to be constant) the bullet experienced while it traveled through the bag? Ans 130 N (Hint: Use work energy theorem to calculate work done by frictional force, and use use the definition of work to find the force)

Answer 1

The correct answer to the question is 130.4 N.

CALCULATION;

The mass of the bullet is given as m = 28 gram = 0. 028 kg.

The initial velocity of the bullet u = 55 m/s

The final velocity of the bullet  v = 18 m/s.

The distance covered by the bullet through the sand bag s = 29 cm.

                                                                                                   = 0.29 m

Let the acceleration of the bullet is a .

From equation of kinematics, we know that-

                                         $v^2-u^2=\ 2as$

                                        ⇒ $a=\ \frac{v^2-u^2}{2s}$

                                               $=\ \frac{(18)^2-(55)^2}{2\times 0.29}\ m/s^2$

                                              $=\ -4656.897\ m/s^2$

The negative sign is used due to the fact that force is opposing in nature. Its velocity is decreasing with time.

From Newton's second law of motion, we know that net force on a body is equal to the product of mass with acceleration.

Mathematically F = ma.

Hence, the frictional force exerted on the bullet is calculated as -

                                              F = m × a

                                                 = 0.028 × (-4656.897) N

                                                 = -130.4 N                     [ANS]

Here, N ( newton) stands for the unit of force.