A block of mass 2.5 kg is sitting on a frictionless ramp with a spring at the bo

Question

A block of mass 2.5 kg is sitting on a frictionless ramp with a spring at the bottom, as shown, that has a spring constant of 475 N/m. The angle of ramp with respect to the horizontal is 36 degrees.

Randomized Variables

m=2.5 kg

k= 475 N/m

Theta = 36 degrees

A) The block, starting at rest, slides down the ramp a distance 52 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 beters 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

given:

m=2.5 kg

k= 475 N/m

Theta = 36 degrees

a) Potential energy gets converted to spring energy. If the spring compresses "x," then

mg(x + 0.52)sin = ½kx²
2.5* 9.8 * (x + 0.52) * sin36 =0.5* 475 * x^2

14.40*x+ 7.488 =237.5*x^2
which is quadratic in x and solves to x = -0.14 m not possible
and x = 0.21 m = 21 cm

b) Spring energy becomes PE and KE:
½kx² = mgxsin + ½mv²
½ * 475N/m * (0.21m)² = 2.5kg * 9.8m/s² * 0.21m * sin36º + ½ * 2.5kg * v²
10.47J = 3.024J + 1.25kg*v²
v = 2.44 m/s

c) mgh = 2.5kg * 9.8m/s² *( -(0.21m + 0.52m) )* sin36º = -10.47 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 = 0.21 m.

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