Physics help please?

RealPro

I also used square grid notebooks once, then I got my sh!t together. Assuming indeed m=20 g (and not 22) It seems right even though you should fix your diagram lol, it's some kind of different story. For maximum mathy goodness you should not do the calculations until the very end, otherwise intermediate steps carry errors, and it looks cleaner... or is it just me? t = sqrt(2h/g) v = d/t = d * sqrt(g/(2h)) mv^2 = kx^2 m d^2 * g/(2h) = kx^2 k = mgd^2 / (2hx^2) = 30.98 N/m = 31 N/m to 2 sig digits

Some Body

y = y₀ + v₀ t + ½ gt² 0 = 1.7 + (0) t + ½ (-9.8) t² 4.9 t² = 1.7 t = 0.589 s x = x₀ + v₀ t + ½ at² 5.1 = 0 + v (0.589) + ½ (0) (0.589)² v = 8.66 m/s 1/2 m v² = 1/2 k x² m v² = k x² (0.022) (8.66)² = k (0.22)² k = 34.1 N/m

electron1

The work that is done by the spring is equal to the kinetic energy of the ball. Work = ½ * k * d^2 = k * 0.0242 The ball moves 5.1 meters horizontally in the same time that it falls 1.7 meters. Let’s use this distance in the following equation to determine the time. d = vi * t + ½ * a * t^2 Since the ball is moving horizontally, its initial vertical velocity is 0 m/s. 1.7 = ½ * 9.8 * t^2 t = √(3.4/9.8) The time is approximately 0.59 second. Let’s use the time and the horizontal distance to determine the ball’s velocity. v = d ÷ t = 5.1 ÷ √(3.4/9.8) The velocity is approximately 8.7 m/s. KE = ½ * 0.022 * [(5.1 ÷ √(3.4/9.8)]^2 = 0.82467 J k * 0.0242 = 0.82467 k = 0.82467 ÷ 0.0242 Rounded to two significant digits, the spring constant is 34 N/m. I hope this is helpful for you.