block of mass 3.5 kg slides down frictionless incline that is 10 m long and inclined at angle of 37 when reaches bottom of incline it collides perfectly inelastically with another block of mass 2 kg then slides along a level floor if coefficient of friction between the block and the floor is .35 how far will it travel?
To determine how far the block will travel along the level floor after colliding with another block, we need to analyze the conservation of momentum during the collision and then consider the subsequent motion on the level floor.
First, let's determine the initial velocities of the blocks before the collision. Since the incline is frictionless, the block of mass 3.5 kg will only experience gravitational force acting along the incline. We can decompose this force into two components: one parallel to the incline and another perpendicular to the incline.
The force parallel to the incline is given by: F_parallel = m * g * sin(theta)
where m is the mass of the block and theta is the angle of the incline.
Substituting the given values into the equation, we have:
F_parallel = 3.5 kg * 9.8 m/s^2 * sin(37°)
F_parallel ≈ 20.69 N
As there is no friction, this parallel force will provide the block’s acceleration down the incline:
a = F_parallel / m
a = 20.69 N / 3.5 kg
a ≈ 5.91 m/s^2
Using the kinematic equation v^2 = u^2 + 2as (where u is the initial velocity, a is the acceleration, and s is the distance), we can find the velocity of the block at the bottom of the incline.
At the top of the incline, the block starts from rest and has no initial velocity (u = 0). The distance covered (s) is given as 10 m.
v^2 = 0^2 + 2 * 5.91 m/s^2 * 10 m
v^2 ≈ 118.2 m^2/s^2
v ≈ √118.2 m^2/s^2
v ≈ 10.87 m/s
Now, let's consider the collision of the two blocks. Since they collide perfectly inelastically, they stick together after the collision.
Using the principle of conservation of momentum:
(mass1 * velocity1) + (mass2 * velocity2) = (mass1 + mass2) * final_velocity
(3.5 kg * 10.87 m/s) + (2 kg * 0) = (3.5 kg + 2 kg) * final_velocity
final_velocity = (3.5 kg * 10.87 m/s) / (3.5 kg + 2 kg)
final_velocity ≈ 6.02 m/s
After the collision, the combined mass of the blocks is 3.5 kg + 2 kg = 5.5 kg. Now, the system is on a level floor with a coefficient of friction of 0.35.
Using the equation F_friction = μ * N (where F_friction is the frictional force, μ is the coefficient of friction, and N is the normal force), we can calculate the frictional force acting on the block.
The normal force N is the product of the mass and the acceleration due to gravity: N = (3.5 kg + 2 kg) * 9.8 m/s^2
F_friction = 0.35 * [(3.5 kg + 2 kg) * 9.8 m/s^2]
F_friction ≈ 13.41 N
Now, we know the frictional force acting on the block. To find the acceleration on the level floor, we'll use the equation F_net = m * a (where F_net is the net force and m is the mass).
Since there is no applied force other than friction, the net force is equal to the frictional force acting on the block, so:
F_net = F_friction
m * a = F_friction
a = F_friction / m
a ≈ 13.41 N / 5.5 kg
a ≈ 2.44 m/s^2
Finally, let's determine the distance covered by the block while decelerating with an acceleration of 2.44 m/s^2.
Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (6.02 m/s), a is the acceleration (-2.44 m/s^2), and s is the distance:
0^2 = 6.02^2 + 2 * (-2.44 m/s^2) * s
0 = 36.24 - 4.88s
4.88s = 36.24
s ≈ 7.44 m
Therefore, the block will travel approximately 7.44 meters on the level floor after colliding with the other block.