A rifle bullet with a mass of 17.5 g traveling toward the right at 293 m/s strikes a large bag of sand and penetrates it to a depth of 23.6 cm. Determine the magnitude and direction of the friction force (assumed constant) that acts on the bullet.
a = (V^2-Vo^2)/2d.
a=(0-(293)^2)/0.472m = -5.5*10^6 m/s^2.
Ff = a/g = -5.5*10^6/9.8 = -5.6*10^5 N.
To the left.
Correction:
a/g = Coefficient of friction,u.
u = a/g = = -5.5*10^6/9.8 = -5.6*10^5.
Ff = u*m*g = -5.6*10^5*0.0175*9.8 = -9.6*10^4 N. To the Left.
Correction:
a = (V^2-Vo^2)/2d.
a = (0-(293)^2)/0.472m = -181,883 m/s^2.
u = a/g = -181883/9.8 = -18,560.
Ff = u*mg = 18560*0.0175*9.8 = 3183 N. To the left.
To determine the magnitude and direction of the friction force acting on the bullet, we first need to calculate the deceleration experienced by the bullet as it penetrates the sand.
First, let's convert the mass of the bullet to kilograms:
Mass of the bullet = 17.5 g = 0.0175 kg
Next, we'll calculate the initial velocity of the bullet:
Initial velocity of the bullet = 293 m/s (given)
Using the equation of motion, which relates displacement, initial velocity, final velocity, acceleration, and time, we have:
Final velocity of the bullet = 0 m/s (since the bullet comes to rest)
We need to find the deceleration, which is the negative acceleration experienced by the bullet. We can use the equation:
Final velocity^2 = Initial velocity^2 + 2 * acceleration * displacement
Rearranging the equation, we get:
Deceleration = (Final velocity^2 - Initial velocity^2) / (2 * displacement)
Deceleration = (0 - 293^2) / (2 * 0.236 m)
Now, we can solve for the deceleration:
Deceleration = -115226 / 0.472 = -244296.61 m/s^2
The deceleration is negative because it is opposite to the direction of the bullet's initial velocity.
The friction force acting on the bullet is equal to the mass of the bullet multiplied by its deceleration, and the direction of the friction force is opposite to the bullet's initial velocity.
Friction force = mass * deceleration
Friction force = 0.0175 kg * (-244296.61 m/s^2)
Friction force = -4274.92 N
Therefore, the magnitude of the friction force acting on the bullet is approximately 4274.92 N, and its direction is opposite to the initial velocity (toward the left).