Question

If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and the spring is uncompressed. The contact surface is smooth.

Answer 1

Answer:

The velocity is  $4.6 m/s^2$

Explanation:

Given:

Force = 500N

Distance  s= 0

To find :

Its velocity at s = 0.5 m

Solution:

$\sum F_{x}=m a$

$F\left(\frac{4}{5}\right)-F_{S}=13 a$

$500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a$

$400-(500 s)=13 a$

$a = \frac{400 -(500s)}{13}$

$a = (30.77 -38.46s) m/s^2$

Using the relation,

$a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}$

$a=v \frac{d v}{d s}$

$v d v=a d s$

Now integrating on both sides

$\int_{0}^{v} v d v=\int_{0}^{0.5} a d s$

$\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s$

$\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}$

$\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right]$

$\left[\frac{v^{2}}{2}\right]=[15.385-4.807]$

$\left[\frac{v^{2}}{2}\right]=10.578$

$v^{2}=10.578 \times 2$

$v^{2}=21.15$

$v = \sqrt{21.15}$

$v = 4.6 m/s^2$