The block shown in (Figure 1) has mass m = 7.0 kg and lies on a fixed smooth frictionless plane tilted at an angle θ = 22.0 ∘ to the horizontal. If the block starts from rest 15.2 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?

Question

The block shown in (Figure 1) has mass m = 7.0 kg and lies on a fixed smooth frictionless plane tilted at an angle θ = 22.0 ∘ to the horizontal. If the block starts from rest 15.2 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?

Answer 1

Why did the block never make it to the party?

Because it got so "slope-y" on the inclined plane and decided to just slide down to the bottom instead!

Now, let's calculate its speed, shall we?

Answer 2

To solve this problem, we can use the principles of conservation of mechanical energy and the concept of work and energy. Here's a step-by-step explanation on how to approach this problem:

Step 1: Identify the given information:
- Mass of the block (m) = 7.0 kg
- Angle of the incline (θ) = 22.0 degrees
- Initial position of the block (s) = 15.2 m
- Acceleration due to gravity (g) = 9.8 m/s^2 (approximate value)

Step 2: Determine the height difference:
The height difference (h) between the initial and final positions of the block can be calculated using trigonometry. Since the block is on a smooth inclined plane, the change in height (h) can be found using the equation: h = s * sin(θ), where s is the initial position.

h = 15.2 m * sin(22.0°)
h ≈ 5.9 m

Step 3: Calculate the potential energy:
The potential energy of the block at the top of the incline is given by the formula: PE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height difference.

PE = 7.0 kg * 9.8 m/s^2 * 5.9 m
PE ≈ 406.58 J

Step 4: Determine the final velocity using conservation of energy:
According to the law of conservation of mechanical energy, the total mechanical energy of the system remains constant. Initially, the block has only potential energy, and at the bottom of the incline, it will have both potential and kinetic energy. Therefore, we can equate the initial potential energy to the final mechanical energy:

PE = KE
m * g * h = 1/2 * m * v^2, where v is the final velocity of the block

Simplifying the equation and solving for v:

v^2 = 2 * g * h
v = √(2 * g * h)

Substituting the known values:

v = √(2 * 9.8 m/s^2 * 5.9 m)
v ≈ 13.2 m/s

Therefore, the block's speed when it reaches the bottom of the incline is approximately 13.2 m/s.

Answer 3

h = 15.2*sin22 = 5.69 m.

V^2=Vo^2 + 2g*h = 0 + 19.6*5.69 = 111.6
V = 10.6 m/s.

The block shown in (Figure 1) has mass m = 7.0 kg and lies on a fixed smooth frictionless plane tilted at an angle θ = 22.0 ∘ to the horizontal. If the block starts from rest 15.2 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?

V =106m/s

Why did the block never make it to the party?

To solve this problem, we can use the principles of conservation of mechanical energy and the concept of work and energy. Here's a step-by-step explanation on how to approach this problem:

h = 15.2*sin22 = 5.69 m.