A parcel is launched at an initial speed of 3.0ms^-1 up a rough plane inclined at an angle of 35 above the horizontal.The coefficient of kinetic energy between the parcel and the plane is 0.30.Determine the maximum distance travelled by the parcel up the plane.
Why did the parcel decide to go up the plane? Because it heard there was a great view from the top! Let's calculate how far it traveled.
Given:
Initial speed (u) = 3.0 m/s
Angle of incline (θ) = 35 degrees
Coefficient of kinetic energy (μ) = 0.30
To find the maximum distance traveled by the parcel, we need to find the height it reaches on the incline.
First, let's break down the initial velocity into horizontal and vertical components. The vertical component (u_y) can be calculated using the formula: u_y = u * sin(θ)
u_y = 3.0 m/s * sin(35°)
u_y ≈ 1.71 m/s
Now, let's find the time it takes for the parcel to reach its maximum height. We can use the vertical component of the initial velocity and the acceleration due to gravity (g = 9.8 m/s^2). The formula to calculate time (t) is: t = u_y / g
t = 1.71 m/s / 9.8 m/s^2
t ≈ 0.17 s
Next, let's calculate the distance covered horizontally during this time. The horizontal distance (d) can be derived using the formula: d = u * cos(θ) * t
d = 3.0 m/s * cos(35°) * 0.17 s
d ≈ 0.53 m
So, the maximum distance traveled by the parcel up the plane is approximately 0.53 meters. It might not seem like much, but from up there, the parcel definitely had a "peak" experience!
To determine the maximum distance travelled by the parcel up the plane, we need to analyze the forces acting on it and apply some fundamental principles of physics.
First, let's consider the forces acting on the parcel. We have the weight force (mg), which acts vertically downwards, and the friction force (Ff) acting parallel to the inclined plane. The friction force can be determined using the coefficient of kinetic friction (μk) and the normal force (N).
The normal force (N) can be determined using the weight force and the angle of inclination (θ):
N = mg cos(θ)
The friction force (Ff) can be determined using the normal force and the coefficient of kinetic friction:
Ff = μk * N
Since the parcel is launched upward, we need to consider the component of gravitational force parallel to the inclined plane. This force can be determined using the weight force and the angle of inclination:
Fg = mg sin(θ)
Now, let's analyze the forces acting on the parcel at the maximum distance traveled. At this point, the friction force will be equal to the force component due to gravity along the plane. So, we can set up the equation:
Ff = Fg
μk * N = mg sin(θ)
Substituting the value of N from the previous equation, we have:
μk * mg cos(θ) = mg sin(θ)
Dividing both sides by mg:
μk * cos(θ) = sin(θ)
Now, we can rearrange this equation to solve for θ:
tan(θ) = μk
Using the given coefficient of kinetic friction (μk = 0.30), we can solve for θ:
θ = tan^(-1)(μk)
θ = tan^(-1)(0.30)
θ ≈ 16.7 degrees
Now, we can use trigonometry to determine the maximum distance traveled by the parcel. We can use the horizontal component of the initial velocity (3.0 m/s) and the angle of inclination (θ).
The horizontal component of the initial velocity is given by:
Vx = 3.0 m/s * cos(θ)
Substituting the value of θ, we have:
Vx = 3.0 m/s * cos(16.7 degrees)
Vx ≈ 2.8 m/s
Now, we can determine the time taken (t) for the parcel to reach its maximum height by using the vertical component of the initial velocity (3.0 m/s) and the angle of inclination (θ):
Vy = 3.0 m/s * sin(θ)
Substituting the value of θ, we have:
Vy = 3.0 m/s * sin(16.7 degrees)
Vy ≈ 0.9 m/s
Using the equation of motion for vertical motion:
Vy = gt - (α * t^2) / 2
Where g is the acceleration due to gravity and α is the component of acceleration along the inclined plane (α = g sin(θ)).
Substituting the values:
0.9 m/s ≈ (9.8 m/s^2) * t - ((9.8 m/s^2) * sin(16.7 degrees) * t^2) / 2
Now, we can solve this quadratic equation for t. The positive root will give us the required time taken to reach the maximum height.
Solving this equation will give us t ≈ 0.38 s.
Now, we can find the maximum distance traveled using the horizontal component of the initial velocity:
Maximum distance = Vx * t
Maximum distance ≈ 2.8 m/s * 0.38 s
Maximum distance ≈ 1.06 meters
Therefore, the maximum distance traveled by the parcel up the plane is approximately 1.06 meters.