3. A box is given an initial speed of 3.0 m/s up an incline where the angle θ =22o. Use the diagram below to define the x-y axes. Assume the original position is on the ground (xo and yo = 0). Ignore friction.
a. How far up the plane (in meters) will the box go?
b. How much total time (in seconds) elapses (starting at
time = 0) before the box returns to its original position?
KE=PE
mv₀²/2=mgh=mgxsinα
x= v₀²/2gsinα=3²/2•9.8•sin22=1.23 m.
a₁=v₀²/2x =9/2•1.23=3.66 m/s²
Upward
v= v₀-a₁t₁
v=0
t₁=v₀/a₁=3/3.66=0.82 s.
Downward
PE=KE
mgxsinα= mv²/2
=> v=v₀ =>
a₂=a₁
=> t₂=t₁
t=2t₁=2•0.82 =1.64 s
To find the solutions to these questions, we need to analyze the motion of the box up the inclined plane. We can break the motion into two components: the motion parallel to the incline (x-direction) and the motion perpendicular to the incline (y-direction).
a. To find how far up the plane the box will go (displacement in the y-direction), we can use the equation:
y = y0 + v0y t + (1/2) a y t^2
In this case, the box starts from rest in the y-direction, so v0y = 0. Also, since the only force acting in the y-direction is gravity pulling the box down, we know that a y = -g, where g is the acceleration due to gravity. The equation becomes:
y = (1/2)(-g)t^2
Now, we need to relate the time t to the motion in the x-direction. To do this, we can use the initial velocity given for the box (v0x = 3.0 m/s) and the angle θ of the incline. The acceleration in the x-direction is given by:
a x = g sin(θ)
Using the equation of motion in the x-direction:
x = x0 + v0x t + (1/2) a x t^2
Since the box starts from rest in the x-direction, we have v0x = 0. The equation simplifies to:
x = (1/2) a x t^2
Now, we can relate the time t to the displacement in the y-direction by using the relationship between x and y:
x = y/tan(θ)
By substituting this expression in the equation for x, we can solve for t:
(1/2) a x t^2 = y/tan(θ)
Substituting the values for g and θ, we have:
(1/2) (g sin(θ)) t^2 = y/tan(θ)
Simplifying, we get:
t = √(2y/gsin(θ))
Now, we can substitute this value of t back into the equation for y:
y = (1/2)(-g)(√(2y/g sin(θ)))^2
Solving for y, we get:
y = gsin(θ)(√(2y/g sin(θ)))^2
Simplifying, we find:
y = (g sin(θ))^2/(2g)
Now, we plug in the values of g and θ to get the displacement y.
b. To find the time it takes for the box to return to its original position, we need to find the time it takes for the displacement in the x-direction to become zero (i.e., the time it takes for the box to move a distance x, stop, and then return to its original position).
Using the equation for x:
x = (1/2) a x t^2
Since the box starts from rest in the x-direction, we have v0x = 0. The equation simplifies to:
x = (1/2) a x t^2
Substituting the value for a x from before:
x = (1/2) (g sin(θ)) t^2
Now, we can solve for t:
t = √(2x/(g sin(θ)))
Substituting the given value of x, we can find the time it takes for the box to return to its original position.
Note: Remember to convert the angle θ from degrees to radians before using it in trigonometric functions.