- #1
physstudent1
- 270
- 1
Hello, the question I'm having trouble with is:
"A 8.00kg block of ice, released from rest at the top of a 1.5 m long frictionless ramp, slides downhill, reaching a speed of 2.5 m/s at the bottom.
What is the angle between the ramp and the horizontal?"
My first attempt to solve this problem was that I would use kinematics to get the acceleration. And then use the acceleration and Newton's laws to find the angle. By using kinematics I got the equation 2.5^2 = 0 + 2(a)(1.5)
I then got a to be 1.64. Using that I set the equation 8gsin(theta) = 1.64(8). Eventually getting theta to equal 9.6 degrees. But when I compared my answer to the a fellow classmate He got 12.275, but I can't really follow his work and there's no answer in the back of the book, so can anyone help?
"A 8.00kg block of ice, released from rest at the top of a 1.5 m long frictionless ramp, slides downhill, reaching a speed of 2.5 m/s at the bottom.
What is the angle between the ramp and the horizontal?"
My first attempt to solve this problem was that I would use kinematics to get the acceleration. And then use the acceleration and Newton's laws to find the angle. By using kinematics I got the equation 2.5^2 = 0 + 2(a)(1.5)
I then got a to be 1.64. Using that I set the equation 8gsin(theta) = 1.64(8). Eventually getting theta to equal 9.6 degrees. But when I compared my answer to the a fellow classmate He got 12.275, but I can't really follow his work and there's no answer in the back of the book, so can anyone help?