)A block with mass m = 11.3 kg slides down an inclined plane of slope angle 43.4 ° with a constant velocity. It is then projected up the same plane with an initial speed 1.45 m/s. How far up the incline will the block move before coming to rest?
To find how far up the incline the block will move before coming to rest, we need to consider the forces acting on the block.
When the block is sliding down the inclined plane with a constant velocity (meaning no acceleration), the force of gravity acting down the incline is balanced by the friction force acting up the incline.
The force of gravity acting down the incline can be calculated using the formula:
F_gravity = m * g * sin(θ)
where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the incline.
The friction force can be calculated using the formula:
F_friction = m * g * cos(θ)
where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the incline.
Since the block is moving with a constant velocity, we can set these two forces equal to each other:
m * g * sin(θ) = m * g * cos(θ)
The mass of the block and the acceleration due to gravity are common to both sides of the equation, so they cancel out:
sin(θ) = cos(θ)
Using trigonometric identity, we have:
tan(θ) = 1
θ = 45°
Now that we know the angle of the incline is 45°, we can calculate the distance up the incline the block will move before coming to rest.
Using the formula for the distance traveled up an incline by a projectile:
d = (v^2 * sin(2θ)) / g
where v is the initial velocity of the block and g is the acceleration due to gravity, we can plug in the given values:
v = 1.45 m/s
θ = 45°
g = 9.8 m/s^2
d = (1.45^2 * sin(2 * 45°)) / 9.8
d = (2.1025 * 0.866) / 9.8
d = 0.87196 / 9.8
d ≈ 0.089 m
Therefore, the block will move approximately 0.089 meters up the incline before coming to rest.