Question

2- a bullet of mass 52 grams is fired horizontally into a 5.2 kg wooden block on a horizontal surface. the coefficient of kinetic friction between the block and the surface is 0.52. the bullet stops in the block which slides straight ahead for 5.2 meter. at what speed is the bullet fired?(apply conservation of energy with friction.)

Answer 1

The speed of the bullet fired is 61.5 J

The initial momentum of the bullet is given by:

p = mv

where m is the mass of the bullet and v is its initial velocity. In this case, m = 0.052 kg and v is the unknown quantity.

Since the bullet stops in the block, the final momentum of the bullet and block is equal to the initial momentum of the bullet. Therefore,

`p = (m_(bullet) + m_(block))v_f`

where m_bullet is the mass of the bullet, m_block is the mass of the block, and v_f is the final velocity of the bullet and block combination. In this case, m_bullet = 0.052 kg, m_block = 5.2 kg, and v_f is the unknown quantity.

The work done by friction is given by:

W_friction = f × d

where f is the force of friction and d is the distance traveled by the bullet and block. In this case, f = μ × N, where μ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the block and bullet, which is given by:

N = (m_bullet + m_block)g

where g is the acceleration due to gravity. In this case, μ = 0.52 and g = 9.8 m/s². The distance traveled by the bullet and block is d = 5.2 m. Therefore,

W_friction = μ × (m_bullet + m_block)g × d

= 0.52 × (0.052 kg + 5.2 kg) × 9.8 m/s² × 5.2 m

≈ 61.5 J