Answer:
a=g(sinθ-μkcosθ)
Explanation:
In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.
wx=m*g*sinθ
wy=m*g*cosθ
As the objects moves down an incline, acceleration in y axis is 0.
Then, by second Newton's Law:
Fy = m*ay
FN - m*g cos θ = 0,
FN=m*g cos θ
In x axis the forces that interacs are the x component of weight and friction force:
Fx = m*ax
mg sen u-FN*μk=m*a
Being friction force, Fr=FN*μk, we replace with its value in below formula:
m*g *sinθ-(m*g*cosθ*μk)=m*a
Then, isolating a:
a=(m*g sinθ-(m*g*cosθ*μk))/m
Solving, we have next equation:
a=g sinθ-(g*cosθ*μk)
Applying distributive property we have:
a=g*(sinθ-μk*cosθ)
<span>Assuming that the momenta of the two pieces are equal: when they have equal velocities, then
the masses of the two pieces are also equal.
Since there is no force from outside of the system, the center of mass moves on with the same velocity as before the equation. So the two pieces must fly at the side side of the mass center, i.e., they must always be at 90° to the side of the mass center. Otherwise it would not be the mass center, respectively the pieces would not have equal velocities.
This is only possible, when the angle of their velocity with the initial direction is 60°.
Because, cos (60°) = 1/2 = v/(2v).</span>
The equation that relates distance, velocities, acceleration, and time is,
d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and
g is the acceleration due to gravity (equal to 9.8 m/s²)
(1) Dropped rock,
(3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s
(2) Thrown rock with V₀ = 26 m/s
(3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s
The difference between the tim,
difference = 24.73 s - 5.61 s
difference = 19.12 s
<em>ANSWER: 19.12 s</em>
