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

The velocity is
`4.6 m/s^2`

Step-by-step explanation:

Given:

Force = 500N

Distance s= 0

To find :

Its velocity at s = 0.5 m

Solution:

`\sum F_(x)=m a`

`F\left((4)/(5)\right)-F_(S)=13 a`

`500\left((4)/(5)\right)-\left(k_(s)\right)=13 a`

`400-(500 s)=13 a`

`a = (400 -(500s))/(13)`

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

Using the relation,

`a=(d v)/(d t)=(d v)/(d s) * (d s)/(d t)`

`a=v (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[(v^(2))/(2)\right]_(0)^(2)=\left[\left(30.77 s-19.23 s^(2)\right)\right]_(0)^(0.5)`

`\left[(v^(2))/(2)\right]=\left[\left(30.77(0.5)-19.23(0.5)^(2)\right)\right]`

`\left[(v^(2))/(2)\right]=[15.385-4.807]`

`\left[(v^(2))/(2)\right]=10.578`

`v^(2)=10.578 * 2`

`v^(2)=21.15`

`v = √(21.15)`

`v = 4.6 m/s^2`