Answer:
(a) [tex] W_{p} = 0.23 Nm [/tex]
(b) [tex] W_{e} = 1.82 Nm [/tex]
(c) v= 4.42 m/s
(d) x = 20cm
Explanation:
(a) The work done on the block by the gravitational force is:
[tex] W_{p} = E_{p} = mgx [/tex]
where [tex]W_{p}[/tex]: is the work by the gravitational force, [tex] E_{p} [/tex]: potential energy, m: is the block's mass, g: gravitational acceleration and x: is the distance of compression
[tex] W_{p} = mgx = 0.210 kg \cdot 9.81 \frac{m}{s^{2}} \cdot 0.11 m = 0.23 N\cdot m [/tex]
(b) The work done on the block by the spring is given by:
[tex] W_{e} = E_{e} = \frac{1}{2} k\cdot x^{2} [/tex]
where [tex]W_{e} [/tex]: work by the spring force, [tex] E_{e} [/tex]: potential elastic energy, k: spring constant, x: distance of compression
[tex] W_{e} = \frac{1}{2} 300 \frac{N}{m} \cdot (0.11m)^{2} = 1.82 N\cdot m [/tex]
(c) The speed of the block just before it hits the spring can be calculated by conservation of energy before and after the impact:
[tex] E_{k} = E_{p} + E_{e} [/tex]
[tex] \frac{1}{2} mv^{2} = mgx + \frac{1}{2} k\cdot x^{2} [/tex] (1)
[tex] \frac{1}{2} mv^{2} = 0.23Nm + 1.82Nm [/tex]
[tex] v = \sqrt \frac{2 (0.23Nm + 1.82Nm)}{0.210kg} = 4.42 \frac{m}{s}[/tex]
(d) Using equation (1) we can determine the spring compression when the speed at impact is duplicated:
[tex] \frac{1}{2} m(2v)^{2} = mgx + \frac{1}{2} k\cdot x^{2} [/tex]
[tex] \frac{1}{2} k\cdot x^{2} + mgx - 2mv^{2} = 0 [/tex] (2)
Solving the quadratic equation (2), we have the next spring compression (x):
[tex] x = 0.20m = 20cm [/tex]
So, an increase in the speed at impact will also increase the spring compression.
I hope it helps you!
