1). You do a physics lab experiment pn another planet. A small block is released
ID: 1591649 • Letter: 1
Question
1). You do a physics lab experiment pn another planet. A small block is released from the rest at the top of a long frictionless ramp that is inclined at an angle of 36.9° above the horizontal. You measure that a small block travels a distance 17.0m down the incline in 8.00s. What is the value of g, the acceleration due to gravity on this planet?(I got 0.32m/s^2, but it's wrong)
2). A small block travels up a frictionless incline that is at an angle of 30.0° above the horizontal. The block has speed 4.05 m/s at the bottom of the incline. Assume g= 9.80 m/s^s. How far up the incline (measured parallel to the surface of the incline) does the block travel before it starts to slide back down?
(I got 0.543 m, but it's telling me it's wrong) 1). You do a physics lab experiment pn another planet. A small block is released from the rest at the top of a long frictionless ramp that is inclined at an angle of 36.9° above the horizontal. You measure that a small block travels a distance 17.0m down the incline in 8.00s. What is the value of g, the acceleration due to gravity on this planet?
(I got 0.32m/s^2, but it's wrong)
2). A small block travels up a frictionless incline that is at an angle of 30.0° above the horizontal. The block has speed 4.05 m/s at the bottom of the incline. Assume g= 9.80 m/s^s. How far up the incline (measured parallel to the surface of the incline) does the block travel before it starts to slide back down?
(I got 0.543 m, but it's telling me it's wrong)
(I got 0.32m/s^2, but it's wrong)
2). A small block travels up a frictionless incline that is at an angle of 30.0° above the horizontal. The block has speed 4.05 m/s at the bottom of the incline. Assume g= 9.80 m/s^s. How far up the incline (measured parallel to the surface of the incline) does the block travel before it starts to slide back down?
(I got 0.543 m, but it's telling me it's wrong)
Explanation / Answer
1)
Acceleration down plane = (2d/t^2) = (34/64) = 0.562m/sec^2.
Force = (ma) = 1 x 0.562 = 0.562N.
g = (0.562/sin 36.9) = 0.936m/sec^2.
2)
Since there is no friction, the KE gets converted into PE:
h = v² / 2g = (4.05m/s)² / 19.6m/s² = 0.836 m
Since the angle was 30º, the slope length is twice the height, and x = 1.672 m.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.