- #1

physstudent1

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"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?