a 5kg suitcase slides 2m down a frictionless ramp incline at 45 degress. How fast is it going at the bottom?
V^2 = 2gd = 2 * 9.8 * 2 = 39.2m.
V = sqrt(39.2) = 6.26m/s.
2m*sin(45)= length of fall (LOF)
this needs to be plugged into equation henry posted. The suitcase does not fall 2m unimpeded. It slides down the ramp at 45 degrees. Thus trig is needed to solve.
To determine the speed of the suitcase at the bottom of the ramp, we can use the principles of physics and basic equations.
First, let's understand the forces acting on the suitcase. Since the ramp is frictionless, the only force acting on the suitcase is gravity. The weight of the suitcase can be calculated using the equation:
Weight = mass × acceleration due to gravity
Given that the mass of the suitcase is 5 kg and acceleration due to gravity is 9.8 m/s^2, the weight of the suitcase is:
Weight = 5 kg × 9.8 m/s^2 = 49 N
Next, we need to determine the component of the weight force acting parallel to the incline. Since the ramp is at a 45-degree angle, the weight can be broken down into two components: one perpendicular to the incline and one parallel to the incline. The parallel component is given by:
Parallel force = weight × sin(angle of incline)
In this case, the angle of incline is 45 degrees. So, the parallel force acting on the suitcase is:
Parallel force = 49 N × sin(45°) ≈ 34.65 N
The parallel force acting on the suitcase is equivalent to the force causing the acceleration of the suitcase down the ramp.
Next, we can use the equation relating force, mass, and acceleration:
Force = mass × acceleration
In this case, the force causing the suitcase to accelerate is the parallel force. So, we can write:
34.65 N = 5 kg × acceleration
Rearranging the equation, we can find the acceleration:
acceleration = 34.65 N / 5 kg ≈ 6.93 m/s^2
Now, we have the acceleration of the suitcase as it slides down the ramp. Finally, we can use the equations of motion to find the final velocity.
The equation linking initial velocity (assuming it starts from rest), acceleration, and distance is:
Final velocity^2 = Initial velocity^2 + 2 × acceleration × distance
Since the initial velocity is 0 (assuming the suitcase starts from rest), the equation becomes:
Final velocity^2 = 2 × acceleration × distance
Plugging in the values for acceleration (6.93 m/s^2) and distance (2 m):
Final velocity^2 = 2 × 6.93 m/s^2 × 2 m = 27.72 m^2/s^2
Taking the square root of both sides to solve for the final velocity:
Final velocity ≈ √(27.72 m^2/s^2) ≈ 5.27 m/s
Therefore, the suitcase is going approximately 5.27 m/s at the bottom of the ramp.