block of mass 5.8 kg is sitting on a frictionless ramp with a spring at the bott
ID: 1487669 • Letter: B
Question
block of mass 5.8 kg is sitting on a frictionless ramp with a spring at the bottom, as shown, that has a spring constant of 560 N/m. The angle of ramp with respect to the horizontal is 17°. A) The block, starting at rest, slides down the ramp a distance 79 cm before hitting the spring. How far, in centimeters, does the spring compress before the block stops? B)After the block comes to rest, the spring pushes the block back up the ramp. How fast, in meters per second, is the block moving right after it comes off the spring? C)What is the difference, in Joules, in gravitational potential energy between the original position of the block at the top of the ramp and the position of the block when the spring is fully compressed?
Explanation / Answer
a) Potential energy gets converted to spring energy. If the spring compresses "x," then
mg(x + 0.31m)sin = 1/2 kx²
5.8 kg * 9.8m/s² * (x + 0.79m) * sin17º = ½ * 560N/m * x²
16.6x +13.1= 280x2
280x2 -16.6x -13.1=0
which is quadratic in x and solves to x = -0.188 m not possible
and x = 0.25m = 25cm
so spring compresses by 25cm
b) Spring energy becomes PE and KE:
½kx² = mgxsin + ½mv²
½ * 560 N/m * (0.25m)² = 5.8kg * 9.8m/s² * 0.25m * sin17º + ½ * 5.8kg * v²
17.5 J = 4.15 J + 2.9kg*v²
v = (13.35J / 2.9kg) = 2.15 m/s
c) mgh = 5.8 kg * 9.8m/s² * -(0.25m + 0.79m) * sin17 = -17.28 J
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