Answers
Answer:
x = 10.75 m
Explanation:
For this problem we will solve it in two parts, the first using energy and the second with kinematics
Let's use the energy work relationship to find the velocity of the block as it exits the ramp
W = [tex]Em_{f}[/tex] - Em₀
Starting point. Higher
Em₀ = U = m g h
the height from the edge of the ramp of the graph has a value
h = 9-3 = 6 m
Final point. At the bottom of the ramp
Em_{f} = K = ½ m v²
Friction force work
W = - fr d
The friction force has the formula
fr = μ N
On the ramp, we can use Newton's second law
N - W cos θ = 0
N = W cos θ
where the angle is obtained from the graph
tan θ = (9-3) / (0.5-4) = -6 / 3.5
θ = tan⁻¹ (-1,714)
θ = -59.7º
the distance d is
d = √ (Δx² + Δy²)
d = √ [(0.5-4)² + (9-3)²]
d = 6.95 m
for which the work is
W = - μ mg cos 59.7 d
we substitute
W = Em_{f} -Em₀
- μ mg cos 59.7 d = ½ m v² - m g h
In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2
- μ g cos 59.7 d = ½ v² - g h
v² = 2g (h - very d coss 59.7)
let's calculate
v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)
v = √ 103.8546
v = 10.19 m / s
in the same direction as the ramp
in the second part we use projectile launch kinematics
let's look for the components of velocity
v₀ₓ = vo cos -59.7
[tex]v_{oy}[/tex] = vo sin (-59,7)
v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s
v_{oy} = 10.19 if (-59.7) = -8.798 m / s
Let's find the time to get to the floor (y = o)
y = y₀ + v_{oy} t - ½ g t²
to de groph y₀=3 m
0 = 3 - 8.798 t - ½ 9.8 t²
t² - 1.796 t - 0.612 = 0
we solve the quadratic equation
t = [1.796 ±√(1.796² + 4 0.612)] / 2
t = [1,795 ± 2,382] / 2
t₁ = 2.09 s
t₂ = -0.29 s
since time must be a positive quantity the correct value is t = 2.09 s
we calculate the horizontal displacement
x = v₀ₓ t
x = 5.14 2.09
x = 10.75 m
The motion of the box, after it exits the incline is the motion and trajectory
of a projectile.
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor is approximately 1.24613 m.
Reasons:
Mass of the block, m = 3.5 kg
Coefficient of kinetic friction, μ = 1.2
Location of the = 0.5 m from the origin
Required:
Horizontal distance between the block's point of contact with the floor and
the bottom right-hand edge of the incline.
Solution:
Let θ represent the angle the incline make with the horizontal.
The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)
Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L
Rise of the incline = 10 - 3 = 7
Run of the incline = 4
L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]
Let ΔP.E.₁ represent the potential energy transferred to kinetic energy
and work along the incline, we have;
Energy of the block at the bottom of the incline, M.E.₂, is found as follows;
K.E.₂ = mgh - m·g·μ·cos(θ)·L
[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]
v ≈ 6.1456 m/s
The vertical component of the velocity is therefore;
[tex]v_y = v \cdot sin(\theta)[/tex]
[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]
From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;
ΔP.E.₁ = 3.5×9.81×7
3 = 5.33588·t + 0.5×9.81·t²
Factorizing, the above quadratic equation, we get;
The time it takes the block to reach the floor, t ≈ 0.40869 seconds
Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]
The horizontal distance, x = vₓ × t
∴ x = 3.04908 × 0.40869 ≈ 1.08194
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor, x ≈ 1.24613 m.
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