A block of mass 5 kg moves along the x-direction subject to the force 𝐹 = (−20𝑥 + 10) N, with the value of 𝑥 in metre. At time 𝑡 = 0 s, it is at rest at position 𝑥 = 1 m. The position and momentum of the block at 𝑡 = (𝜋/4) s are

The correct option is: 0.5 m, -5 kg m/s

Let's solve the problem step-by-step.

We are given:
Mass of the block, $m=5\text{kg}$
Force acting on the block, $F\left(x\right)=-20x+10$
At time $t=0\text{s}$, the block is at rest, meaning its velocity is $v\left(0\right)=0$, and its position is $x\left(0\right)=1\text{m}$.

We need to find the position and momentum of the block at $t=\frac{\pi }{4}\text{s}$.

First, let's write down the equation of motion using Newton's second law:
$F=ma=m\frac{d^{2}x}{d{t}^2}$
Substituting the given mass and force:
$5\frac{d^{2}x}{d{t}^2}=-20x+10$
Divide the entire equation by 5:
$\frac{d^{2}x}{d{t}^2}=-4x+2$
Rearranging the terms, we get:
$\frac{d^{2}x}{d{t}^2}+4\left(x-0.5\right)=0$

Let us define a new variable for the displacement from the equilibrium position:
$X=x-0.5$
Since 0.5 is a constant, we have:
$\frac{d^{2}X}{d{t}^2}=\frac{d^{2}x}{d{t}^2}$
Thus, the equation of motion becomes:
$\frac{d^{2}X}{d{t}^2}+4X=0$
This is the standard equation for Simple Harmonic Motion (SHM):
$\frac{d^{2}X}{d{t}^2}+{\omega }^{2}X=0$
Comparing the two, the angular frequency is:
${\omega }^2=4\Rightarrow \omega =2\text{rad/s}$

The general solution for the displacement $X\left(t\right)$ is:
$X\left(t\right)=Acos\left(\omega t\right)+Bsin\left(\omega t\right)$
Substituting $X=x-0.5$ and $\omega =2$:
$x\left(t\right)-0.5=Acos\left(2t\right)+Bsin\left(2t\right)$
$x\left(t\right)=0.5+Acos\left(2t\right)+Bsin\left(2t\right)$

Let's apply the initial conditions to find the constants $A$ and $B$.1. At $t=0$, $x\left(0\right)=1\text{m}$:
$1=0.5+Acos\left(0\right)+Bsin\left(0\right)$
$1=0.5+A\Rightarrow A=0.5\text{m}$

2. The velocity $v\left(t\right)$ is the time derivative of $x\left(t\right)$:
$v\left(t\right)=\frac{dx}{dt}=-2Asin\left(2t\right)+2Bcos\left(2t\right)$
At $t=0$, the block is at rest, so $v\left(0\right)=0$:
$0=-2Asin\left(0\right)+2Bcos\left(0\right)$
$0=2B\Rightarrow B=0$

Thus, the position as a function of time is:
$x\left(t\right)=0.5+0.5cos\left(2t\right)$
And the velocity as a function of time is:
$v\left(t\right)=-sin\left(2t\right)$

Now, let's calculate the position and velocity at $t=\frac{\pi }{4}\text{s}$:Position:
$x\left(\frac{\pi }{4}\right)=0.5+0.5cos\left(2\cdot \frac{\pi }{4}\right)$
$x\left(\frac{\pi }{4}\right)=0.5+0.5cos\left(\frac{\pi }{2}\right)$
Since $cos\left(\frac{\pi }{2}\right)=0$:
$x\left(\frac{\pi }{4}\right)=0.5\text{m}$

Velocity:
$v\left(\frac{\pi }{4}\right)=-sin\left(2\cdot \frac{\pi }{4}\right)$
$v\left(\frac{\pi }{4}\right)=-sin\left(\frac{\pi }{2}\right)$
Since $sin\left(\frac{\pi }{2}\right)=1$:
$v\left(\frac{\pi }{4}\right)=-1\text{m/s}$

Momentum:
The momentum $p$ of the block is given by:
$p=m\cdot v$
$p=5\text{kg}\cdot \left(-1\text{m/s}\right)=-5\text{kg m/s}$

Therefore, at $t=\frac{\pi }{4}\text{s}$, the position is 0.5 m and the momentum is -5 kg m/s.