A block of mass 5 kg moves along the -direction subject to the force N, with the value of in metre. At time s, it is at rest at position The position and momentum of the block at s are [2024]

Question

A block of mass 5 kg moves along the $x$ -direction subject to the force $F = (- 20 x + 10)$ 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 $t = (\frac{π}{4})$ s are [2024]

Smart Achievers · Accepted Answer

(3)

Given mass of block = 5kg moving along the x-direction subject to the force $F = (- 20 x + 10)$ N with the value of $x$ in metre.

$Acceleration a = \frac{F}{m} \overset{F = (- 20 x + 10) N}{\leftarrow} m = 5 kg$

$t = 0, v = 0, x = 1 m$

$= \frac{- 20 x + 10}{5} = - 4 x + 2$

$Also, a = v \frac{d v}{d x} = - 4 x + 2$

$∴$ $\int_{0}^{v} v d v = \int_{1}^{x} (- 4 x + 2) d x$ $\Rightarrow \frac{v^{2}}{2} = {(- 2 x^{2} + 2 x)}_{1}^{x}$

$or, v = - 2 \sqrt{x - x^{2}} [since particle starts moving in -ve x-direction]$

$∴$ $\frac{d x}{d t} = - 2 \sqrt{x - x^{2}}$ $\Rightarrow \int_{x = 1}^{x = x} \frac{d x}{\sqrt{x - x^{2}}} = - 2 \int_{0}^{\frac{π}{4}} d t$

$\Rightarrow \sin^{- 1} {(2 x - 1)}_{1}^{x} = - \frac{π}{2}$

$∴$ $Position x = 0.5 m$

$And since v = - 2 \sqrt{x - x^{2}} = - 2 \sqrt{0.5 - {(0.5)}^{2}} = - 1 m/s$

$∴$ $Momentum P = m v = 5 (- 1) = - 5 {kg ms}^{- 1}$

Solution

Q.
A block of mass 5 kg moves along the $x$ -direction subject to the force $F = (- 20 x + 10)$ 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 $t = (\frac{π}{4})$ s are [2024]

1 −0.5 m, 5 kg m/s

2 0.5 m, 0 kg m/s

3 0.5 m, −5 kg m/s

4 −1 m, 5 kg m/s

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