A block with mass m = 19.1 kg slides down an inclined plane of slope angle 18.2 o with a constant velocity. It is then projected up the same plane with an initial speed 2.05 m/s. How far up the incline will the block move before coming to rest?
To find the distance the block will move up the incline before coming to rest, we can use the concept of work and energy.
First, let's consider the block sliding down the inclined plane with a constant velocity. Since the velocity is constant, the net force acting on the block must be zero.
The gravitational force acting on the block can be resolved into two components: one parallel to the incline (mg sinθ) and one perpendicular to the incline (mg cosθ).
Since the net force is zero, the force of friction opposing the block's motion up the incline must be equal in magnitude to the component of the gravitational force parallel to the incline (mg sinθ). Therefore, the force of friction can be expressed as:
Friction force (f) = mg sinθ
Next, let's consider the block being projected up the incline with an initial speed of 2.05 m/s. The block will continue moving up until the force of friction stops it.
The work done by the force of friction can be calculated using the formula:
Work (W) = force x distance x cosθ
Since the block comes to rest, the work done by the force of friction must be equal to the initial kinetic energy of the block (1/2 mv^2):
W = (1/2)mv^2
Setting the work done by friction equal to the initial kinetic energy:
mg sinθ x d x cosθ = (1/2)mv^2
We can cancel out the mass and solve for the distance (d):
d = (1/2) v^2 / (g sinθ x cosθ)
Substituting the given values:
d = (1/2) * 2.05^2 / (9.8 * sin(18.2) * cos(18.2))
Calculating this expression will give us the distance up the incline the block will move before coming to rest.