Answer:
k = 26.25 N/m
Explanation:
given,
mass of the block= 0.450
distance of the block = + 0.240
acceleration = a_x = -14.0 m/s²
velocity = v_x = + 4 m/s
spring force constant (k) = ?
we know,
x = A cos (ωt - ∅).....(1)
v = - ω A cos (ωt - ∅)....(2)
a = ω²A cos (ωt - ∅).........(3)
![\omega = \sqrt{\dfrac{k}{m}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%7D)
now from equation (3)
![a_x = \dfrac{k}{m}x](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cdfrac%7Bk%7D%7Bm%7Dx)
![k = \dfrac{m a_x}{x}](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7Bm%20a_x%7D%7Bx%7D)
![k = \dfrac{0.45 \times (-14)}{0.24}](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7B0.45%20%5Ctimes%20%28-14%29%7D%7B0.24%7D)
k = 26.25 N/m
hence, spring force constant is equal to k = 26.25 N/m
Answer:
The appropriate response is "
". A further explanation is described below.
Explanation:
The torque (
) produced by the force on the dam will be:
⇒ ![d \tau=XdF](https://tex.z-dn.net/?f=d%20%5Ctau%3DXdF)
On applying integration both sides, we get
⇒ ![\tau = \int_{0}^{a}x pgL(h-x)dx](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cint_%7B0%7D%5E%7Ba%7Dx%20pgL%28h-x%29dx)
⇒ ![= pgL\int_{0}^{h}(h-x)dx](https://tex.z-dn.net/?f=%3D%20pgL%5Cint_%7B0%7D%5E%7Bh%7D%28h-x%29dx)
⇒ ![=pgL[\frac{h^3}{2} -\frac{h^3}{3} ]](https://tex.z-dn.net/?f=%3DpgL%5B%5Cfrac%7Bh%5E3%7D%7B2%7D%20-%5Cfrac%7Bh%5E3%7D%7B3%7D%20%5D)
⇒ ![=\frac{1}{6} PgLh^3](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B6%7D%20PgLh%5E3)
The force would be coming from the right causing the box the lean/ slide to left, if it wasnt sticky.