Physics spring problem?
A block of mass 3.9 kg is sitting on a frictionless ramp with a spring at the bottom as shown, that has a spring constant of 575 N/m. The angle of ramp with respect to the horizontal is 39 degrees.
a) The block, starting at rest, slides down the ramp a distance 31cm 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 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?
- NCSLv 75 years agoFavorite Answer
a) Potential energy gets converted to spring energy. If the spring compresses "x," then
mg(x + 0.31m)sinΘ = ½kx²
3.9kg * 9.8m/s² * (x + 0.31m) * sin39º = ½ * 575N/m * x²
which is quadratic in x and solves to x = -0.12 m ← not possible
and x = 0.21 m = 21 cm
b) Spring energy becomes PE and KE:
½kx² = mgxsinΘ + ½mv²
½ * 575N/m * (0.21m)² = 3.9kg * 9.8m/s² * 0.21m * sin39º + ½ * 3.9kg * v²
12.7 J = 5.1J + 1.95kg*v²
v = √(7.6J / 1.95kg) = 2.0 m/s
c) mgΔh = 3.9kg * 9.8m/s² * -(0.21m + 0.31m) * sin39º = -12.5 J
The answer should be the same as the LHS in part (b) (the spring energy) -- the difference is due to rounding in the value of x. Precisely, x = 0.208218 m.
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