A bullet of mass 3.0g moving at 500m/s hits a tree and pierced 20cm into the trunk of the tree.what is the resistimg force offered by the tree?
Force * distance = Work done by bullet = Initial kinetic energy of the bullet
Solve for the force
F = M*V^2/(2*X)
= 3*10^-3*(500)^2/(2*0.2) Newtons
= 1875 N
To find the resisting force offered by the tree, we'll first need to calculate the deceleration of the bullet using the equation of motion. The deceleration is given by the equation:
a = (v^2 - u^2) / (2s)
where:
a = acceleration (in this case, deceleration since the bullet is slowing down)
v = final velocity (0 m/s since the bullet comes to a stop)
u = initial velocity (500 m/s, as given)
s = displacement (20 cm, which we'll convert to meters)
Converting the displacement from centimeters to meters:
20 cm = 0.20 m
Substituting the values into the equation:
a = (0 - 500^2) / (2 * 0.20)
Calculating:
a = (-250,000) / 0.40
a ≈ -625,000 m/s² (deceleration)
Now, we can calculate the resisting force using Newton's second law of motion, which states that force (F) is equal to the mass (m) multiplied by the acceleration (a):
F = m * a
Substituting the values:
F = 0.003 kg * -625,000 m/s²
Calculating:
F ≈ -1,875 N
The negative sign indicates that the force is acting in the opposite direction of the bullet's motion, as it opposes the bullet's movement and causes it to decelerate. So, the resisting force offered by the tree is approximately 1,875 Newtons.