Question

5. A bullet of mass m-0.02 kg was horizontally shot into a wood block of mass mo 8.98 kg, which is attached at one end of a spring with a spring constant k = 100 N/m, as shown in right Figure. After the bullet was embedded into the wood block, the spring was compressed 10 cm. If the coefficient of kinetic friction is 0.2 between the wood block and the surface, find the initial velocity of the bullet. (10 points)

Answer 1

mo F m Giren, mass of the bullet m= 0.02 kg - mass of the wood block, m's 8.98 kg Spring constant of the spring ek = 100 N/m. Compressed distance of spring x= 10cm so-, co-efficient of kinetic friction = 0.2 between block and the surface. Let the initial velocity of the bullet is a v. So, the kinetic energy of the ballet moz $ *0.02 *V2 = 0.01 02 before the bulet was embeded in the block the total chergy out of the system wan, bis 0.0162... as because spring was his stable point But after hitting the bullet it was needed into the block. then total man of the system aM = a moema 8.98 +0.02 -9 kg. 2. Friction force of = ung = 0.2 x 9 x 9.8.N. = 17.64 N. 2. work to overcome this force-fx-1.7645 in this spring s o aby x 100x (0.125 to the potential energy = 0.5, J. from Energy conservation law we get

5 Eo fatt unt= 17.644 0.5 = 18.14 2 0.01 02 = 18.14 18.14 o vzo. of os, v2=1814 av=42.59 m/s. nelocity of the bullet, v= 42.59 m/s.