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=1m. the position and momentum of the block at t=pi/4s are

 Alright, let's tackle this problem step by step. I'm going to break it down to understand how the block moves under the given force and then determine its position and momentum at 

$t=\frac{\pi }{4}$ seconds.

Understanding the Problem

We have a block with:

• Mass $m=5$ kg

• Force acting on it: $F=-20x+10$ N, where $x$ is the position in meters

• Initial conditions:

  • At $t=0$ s: $x=1$ m

  • At $t=0$ s: $v=0$ m/s (since it's at rest)

We need to find:

• The position $x$ at $t=\frac{\pi }{4}$ s

• The momentum $p$ at $t=\frac{\pi }{4}$ s

Analyzing the Force

The force is given by:

$$F=-20x+10$$

This can be rewritten as:

$$F=-20(x-0.5)$$

This resembles the force from a spring (Hooke's law: $F=-kx$), but shifted. Here, the equilibrium position isn't at $x=0$, but at $x=0.5$ m, because that's where $F=0$:

$$-20(x-0.5)=0⟹x=0.5\text{ m}$$

So, the force is restoring towards $x=0.5$ m with a "spring constant" $k=20$ N/m.

Equation of Motion

Newton's second law:

$$F=ma$$$$-20(x-0.5)=5\frac{d^{2}x}{d{t}^2}$$$$\frac{d^{2}x}{d{t}^2}=-4(x-0.5)$$

Let’s define a new variable $y=x-0.5$. Then:

$$\frac{d^{2}y}{d{t}^2}=-4y$$

This is the differential equation for simple harmonic motion (SHM):

$$\frac{d^{2}y}{d{t}^2}+{\omega }^{2}y=0$$

where ${\omega }^2=4$, so $\omega =2$ rad/s.

General Solution for SHM

The general solution for $y(t)$ is:

$$y(t)=A\cos (\omega t)+B\sin (\omega t)$$$$y(t)=A\cos (2t)+B\sin (2t)$$

Then, $x(t)=y(t)+0.5$:

$$x(t)=A\cos (2t)+B\sin (2t)+0.5$$

Applying Initial Conditions

At $t=0$:

1. $x(0)=1$:

$$$$

= A \cos(0) + B \sin(0) + 0.5 ]

$$1=A+0.5$$$$A=0.5$$

2. $v(0)=0$:
   First, find velocity $v(t)$:

$$$$

(t) = \frac{dx}{dt} = -2A \sin(2t) + 2B \cos(2t) ]
At $t=0$:

$$0=-2A\sin (0)+2B\cos (0)$$$$0=0+2B$$$$B=0$$

So, the position as a function of time is:

$$x(t)=0.5\cos (2t)+0.5$$

Finding $x$ at $t=\frac{\pi }{4}$ s

$$x\left(\frac{\pi }{4}\right)=0.5\cos (2\cdot \frac{\pi }{4})+0.5$$$$=0.5\cos \left(\frac{\pi }{2}\right)+0.5$$$$=0.5\cdot 0+0.5$$$$=0.5\text{ m}$$

Finding Momentum at $t=\frac{\pi }{4}$ s

Momentum $p=mv$.

First, find $v(t)$:

$$v(t)=\frac{dx}{dt}=-2\cdot 0.5\sin (2t)$$$$v(t)=-\sin (2t)$$

At $t=\frac{\pi }{4}$:

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

Then, momentum:

$$p=mv=5\cdot (-1)=-5\text{ kg m/s}$$

Verifying the Results

Let me double-check the calculations:

1. Force and SHM:

   • $F=-20(x-0.5)$, so ${\omega }^2=\frac{k}{m}=\frac{20}{5}=4$, $\omega =2$. Correct.

2. Initial conditions:

   • $x(0)=1$: $A=0.5$. Correct.

   • $v(0)=0$: $B=0$. Correct.

3. Position at $t=\frac{\pi }{4}$:

   • $\cos (\frac{\pi }{2})=0$, so $x=0.5$. Correct.

4. Velocity at $t=\frac{\pi }{4}$:

   • $\sin (\frac{\pi }{2})=1$, so $v=-1$. Correct.

   • Momentum $p=-5$. Correct.

Everything checks out.

Final Answers

At $t=\frac{\pi }{4}$ s:

• Position: $x=0.5$ m

• Momentum: $p=-5$ kg m/s

Expressed in the required format:

• Position: $\boxed{{\displaystyle 0.5\text{ m}}}$

• Momentum: $\boxed{{\displaystyle -5\text{ kg m/s}}}$