(a) The spring constant is 1,279.1 N/m.
(b) The length of the ramp is 1.75 m.
(c) The mass did not jump the gap.
(d) The landing area must be placed 1.98 m below the top of the ramp for the block to successfully jump it.
<h3>
Spring constant</h3>
The spring constant is calculated as follows;
¹/₂kx² = ¹/₂mv²
kx² = mv²
k = mv²/x²
k = (5.22 x 4.43²) / (0.283)²
k = 1,279.1 N/m
<h3>Displacement of the mass</h3>
The height traveled by the mass when it reaches top of the ramp is calculated as follows;
¹/₂Kd² = ¹/₂mv² + mgh
The height of the ramp, h = Lsinθ
¹/₂Kd² = ¹/₂mv² + mgLsinθ
0.5 x 1279.1 x 0.283² = 0.5 x 5.22 x (2.81)² + 5.22 x 9.8 x (sin20) x L
51.22 = 20.61 + 17.5L
30.61 = 17.5L
L = 30.61 / 17.5
L = 1.75 m
<h3>Height of the ramp</h3>
h = L x sin20
h = 1.75 x sin20
h = 0.6 m
Time taken for the mass to travel the height
![h = vt + \frac{1}{2} gt^2\\\\0.6 = 2.81t + 4.9t^2\\\\4.9t^2 + 2.81t - 0.6=0\\\\t = 0.17 \ s](https://tex.z-dn.net/?f=h%20%3D%20vt%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5C0.6%20%3D%202.81t%20%2B%204.9t%5E2%5C%5C%5C%5C4.9t%5E2%20%2B%202.81t%20-%200.6%3D0%5C%5C%5C%5Ct%20%3D%200.17%20%5C%20s)
Horizontal distance = vt = 0.17 x 2.81 = 0.48 m <em>(0.48 m is less than 1.16 m)</em>
Thus, the mass did not jump the gap.
<h3>Time of motion if the mass must jump 1.16 m</h3>
t = x/v
t = 1.16/2.81
t = 0.41
<h3>Height it must be placed</h3>
![H = 2.81(0.41) + \frac{1}{2} (9.8) (0.41)^2\\\\H = 1.98 \ m](https://tex.z-dn.net/?f=H%20%3D%202.81%280.41%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%289.8%29%20%280.41%29%5E2%5C%5C%5C%5CH%20%3D%201.98%20%5C%20m)
Thus, the landing area must be placed 1.98 m below the top of the ramp for the block to successfully jump it.
Learn more about conservation of energy here: brainly.com/question/166559