Question

A spring is attached to an inclined plane as shown in the figure. A block of mass m = 2.27 kg is placed on the incline at a distance d = 0.327 m along the incline from the end of the spring. The block is given a quick shove and moves down the incline with an initial speed v = 0.750 m/s. The incline angle is = 20.0°, the spring constant is k = 455 N/m, and we can assume the surface is frictionless. By what distance (in m) is the spring compressed when the block momentarily comes to rest?

Answer 1

Answer:

0.1351 m

Explanation:

*my number calculations may be wrong, but the process is correct*

W = Wg - Ws = mgh - $\frac{1}{2}$kx^2 = mg(d + x)sinθ - $\frac{1}{2}$kx^2

Δk = 0 - $\frac{1}{2\\}$mv$_{i}$^2

W = Δk

-$\frac{1}{2}$mv$_{i}$^2 = mg(d + x)sinθ - $\frac{1}{2}$kx^2

Now plug in your given values:

m=2.27, g=9.8, d=0.327, θ=20, k=455, v=0.750

Now rearrange the equation so it is in the form ax^2 + bx + c = 0 where x is the unknown distance you are looking for.

~if my math is correct it should be:~

0 = -227.5x^2 + 7.608580108x + 3.126443195

Now plug your numbers into the quadratic formula and the positive x value you get is the answer.

~if my math is correct it should be:~

x = 0.1351376847 m = 0.1351 m