a bullet of mass 0.04kg moving with a speed of 90m/s enter a heavy wooden block is stopped after a distance of 60cm.What is the avg resistive force exertes by the block or the bullet
a=(V^2-Vo^2)/2d=(0-90^2)/1.2m=-6750m/s^2
F = m*a = 0.04 * -6750 = -270 N.
To determine the average resistive force exerted by the block on the bullet, we can use the principles of work and energy. The work done by the resistive force would be equal to the change in kinetic energy of the bullet.
First, let's calculate the initial kinetic energy of the bullet using the formula:
Kinetic Energy (KE) = 1/2 * mass * velocity^2
Given:
Mass of the bullet (m) = 0.04 kg
Velocity of the bullet (v) = 90 m/s
KE = 1/2 * 0.04 kg * (90 m/s)^2
KE = 180 J (Joules)
Next, let's calculate the final kinetic energy of the bullet. Since the bullet comes to rest, the final kinetic energy is zero.
Final KE = 0 J
The work done by the resistive force is given by:
Work (W) = Final KE - Initial KE
W = 0 J - 180 J
W = -180 J
The negative sign indicates that the work done by the resistive force is in the direction opposite to the bullet's motion, which makes sense as the bullet is being decelerated.
Finally, we can determine the average resistive force using the equation:
Work (W) = Force (F) * Distance (d)
F = W / d
The distance traveled by the bullet is given as 60 cm, which is 0.6 meters.
F = -180 J / 0.6 m
F ≈ -300 N
Therefore, the average resistive force exerted by the block on the bullet is approximately 300 Newtons (N).