A 50kg block is dropped from a point 10m above as shown in the fig above. if the force constant of the spring is 4.0 x 10\(^4\)Nm-1, find the maximum compression of the spring
[g = 10ms-2]
- 1.25m
-
0.50m
- 0.25m
- 0.05m
The correct answer is: B
Explanation
P.E\(_{mgh}\) = \(\frac{1}{2}\) * k * x\(^2\)
mgh = \(\frac{1}{2}\) * k * x\(^2\)
50 * 10 * 10 = \(\frac{1}{2}\) * 4 * 10\(^4\) * x\(^2\)
→ x\(^2\) = \(\frac{5000 * 2}{4 * 10^4}\)
→ x = √[0.25]
: x = 0.50m