A block of mass 60kg at rest on top of a smooth inclined plane slides down. if the angle and height of the inclined plane are 30degree and 1.0m. calculate the time taken for the block to reach the bottom of the inclined plane.
Please someone should help out
acceleration down the plane is a = g sin30° = 4.9 m/s^2
The length of the plank is 1.0/sin30° = 2.0
so now you need to find t such that 1/2 at^2 = 2.0
To calculate the time taken for the block to reach the bottom of the inclined plane, we can use the concept of acceleration due to gravity and the equations of motion.
First, let's analyze the forces acting on the block:
1. Weight (mg): This is the force pulling the block downwards. Its magnitude can be calculated as follows: Weight = mass × acceleration due to gravity = 60 kg × 9.8 m/s² = 588 N.
2. Normal force (N): This force is perpendicular to the inclined plane and balances the component of weight acting perpendicular to the plane. The magnitude of the normal force can be calculated as follows: N = Weight × cos(θ), where θ is the angle of the inclined plane (30 degrees). Therefore, N = 588 N × cos(30°) = 509.9 N.
3. Force parallel to the inclined plane (f): This is the force responsible for the motion of the block down the incline. Its magnitude can be calculated as follows: f = Weight × sin(θ), where θ is the angle of the inclined plane (30 degrees). Thus, f = 588 N × sin(30°) = 294 N.
Since the inclined plane is smooth, there is no friction to consider in this case.
With these forces in mind, we can now determine the acceleration of the block.
The net force acting on the block is the force parallel to the inclined plane (f). Since there is no net force in the perpendicular direction, the block does not move in that direction. Therefore, the acceleration (a) experienced by the block is given by a = f / mass = 294 N / 60 kg = 4.9 m/s².
Now, let's calculate the distance traveled by the block.
The height of the inclined plane (h) is given as 1.0 m. Since the inclined plane forms a right-angled triangle with the ground, the distance traveled along the incline is equal to the base of this triangle. Using trigonometry, we can determine this distance: base = hypotenuse × sin(θ) = h / sin(θ) = 1.0 m / sin(30°) = 2.0 m.
Now, we have the values for acceleration (a) and distance (d).
Using the second equation of motion: d = (1/2)at², we can rearrange the equation to solve for time (t):
t = √(2d / a) = √(2 × 2.0 m / 4.9 m/s²) ≈ 0.634 s.
Therefore, it will take approximately 0.634 seconds for the block to reach the bottom of the inclined plane.