a bullet moving at 270 m/s penetrates 5 cm into a tree limb, beforecoming to rest assuming that the force exerted by the tree limb is uniform, find its magnitude? mass of the bullet is 10 g
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To find the magnitude of the force exerted by the tree limb, we will first need to calculate the deceleration (negative acceleration) experienced by the bullet. We can then use Newton's second law of motion, F = ma, to find the force.
Let's start by converting the mass of the bullet from grams to kilograms:
mass of bullet = 10 g = 10/1000 kg = 0.01 kg
The initial velocity of the bullet is 270 m/s, and assuming it comes to rest, the final velocity will be 0 m/s.
Therefore, the change in velocity (Δv) can be calculated as:
Δv = final velocity - initial velocity
Δv = 0 m/s - 270 m/s
Δv = -270 m/s
The distance penetrated by the bullet is 5 cm, which is equivalent to 0.05 m:
distance = 5 cm = 5/100 m = 0.05 m
To calculate the deceleration, we can use the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration (or deceleration), and s is the distance traveled.
Rearranging the equation to isolate the deceleration (a) term, we have:
a = (v^2 - u^2) / (2s)
Substituting the values into the equation, we get:
a = (0^2 - (-270)^2) / (2 * 0.05)
Simplifying further, we have:
a = 364500 m^2/s^2 / 0.1
a = 3645000 m^2/s^2
Now we can find the magnitude of the force exerted by the tree limb by using Newton's second law of motion, F = ma:
F = (0.01 kg) * (3645000 m^2/s^2)
F = 36450 N
Therefore, the magnitude of the force exerted by the tree limb is 36450 Newtons.