A cannon fires a shell at an angle of 55.0 above the horizontal. The shell has a mass of 17.5kg, and its initial speed as it leaves the cannon is 289m/s. The cannon is not anchored to the ground. It is mounted on a base -a large block of wood- which is sittin go n a level roadway. If the road exerts a steady kinetic friction force of 430 N on the cannons sliding base, how much time does it take for the cannon to stop sliding after it fires the shell.

the horizontalmomentum of the cannon is 289*cos55 * 17.5

So, force*time=that momentum
solve for time.

To find the time it takes for the cannon to stop sliding after firing the shell, we need to calculate the net force acting on the cannon's sliding base.

First, let's consider the forces acting on the cannon:

1. Gravitational Force (Fg): This force acts vertically downward and has a magnitude equal to the weight of the cannon. We can calculate it using the formula Fg = mg, where m is the mass of the cannon (including the shell) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Fg = (mass of cannon) * g
Fg = 17.5 kg * 9.8 m/s^2
Fg ≈ 171.5 N

2. Normal Force (Fn): This force acts vertically upward and is equal in magnitude and opposite in direction to the gravitational force. Since the cannon is not accelerating vertically, the normal force must balance the weight of the cannon.

Fn = Fg ≈ 171.5 N

Now, let's calculate the net force on the cannon's sliding base:

The net force (Fnet) is the vector sum of all the forces acting on the cannon's sliding base. In this case, we have two horizontal forces: the force of kinetic friction and the horizontal component of the gravitational force.

1. Force of Kinetic Friction (Fk): This force opposes the motion of the cannon and acts horizontally in the opposite direction of the sliding motion. Its magnitude is given as 430 N.

Fk = 430 N

2. Horizontal Component of Gravitational Force (Fgx): This force acts horizontally in the same direction as the sliding motion. We can calculate it using trigonometry, considering that the angle of the cannon above the horizontal is 55.0°.

Fgx = Fg * cos(55.0°)

Fgx = 171.5 N * cos(55.0°)
Fgx ≈ 90.17 N

Now, we can calculate the net force:

Fnet = Fk - Fgx
Fnet = 430 N - 90.17 N
Fnet ≈ 339.83 N

Since the mass of the cannon's sliding base is not given, we cannot calculate its acceleration directly. However, we can determine the force required to stop the cannon using the equation F = ma, where a is the acceleration of the cannon.

F = ma
339.83 N = (mass of sliding base) * a

Now, we know that the acceleration of the cannon's sliding base is equal to the acceleration of the shell fired from the cannon. To find that acceleration, we can use the equations of motion in the horizontal direction.

The horizontal motion of the shell is independent of the vertical motion, so we can treat it as a projectile launched at an angle. The horizontal component of the initial velocity (v₀x) is given by v₀x = v₀ * cos(θ), where v₀ is the initial speed of the shell and θ is the launch angle.

v₀x = 289 m/s * cos(55.0°)
v₀x ≈ 144.09 m/s

We know that the acceleration in the horizontal direction (a) will eventually bring the horizontal velocity to zero. The time it takes for this to happen can be found using the equation:

v = v₀ + at
0 = 144.09 m/s + a * t

From this equation, we can solve for time (t):

t = -144.09 m/s / a

Now, we substitute the force F = 339.83 N and the mass m = 17.5 kg into the equation F = ma:

F = ma
339.83 N = 17.5 kg * a

Now, we can solve for the acceleration (a):

a = 339.83 N / 17.5 kg
a ≈ 19.42 m/s²

Finally, substitute the calculated acceleration (a) into the equation for time (t):

t = -144.09 m/s / a
t ≈ -144.09 m/s / 19.42 m/s²
t ≈ -7.42 s

The negative sign indicates that the time taken for the cannon to stop sliding is 7.42 seconds. However, since time cannot be negative, we take the positive value:

t ≈ 7.42 s

Therefore, it takes approximately 7.42 seconds for the cannon to stop sliding after firing the shell.