The conservative force exerted on a given particle as a function of position x is found to be Fx-3x-1 for positive x values- where x is given in meters-From x-1 m to x-3 m- by how much does the particle-s potential energy change-

The correct option is D 2.079 J The force exerted by weird #160; #160; #160; #160; F #160;= 3x+1 #160;NPotential energy #160; ...

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Standard XII

Physics

Relation between Work Done and PE

The conservat...

Question

The conservative force exerted on a given particle as a function of position $x$ is found to be $F (x) = \frac{- 3}{x + 1}$ for positive $x$ values, where $x$ is given in meters.
From $x = 1 m$ to $x = 3 m$ , by how much does the particle's potential energy change?

0.000 J

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0.693 J

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0.750 J

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2.079 J

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4.159 J

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Solution

The correct option is D $2.079 J$
The force exerted by weird $F = \frac{- 3}{x + 1}$ N

Potential energy $U (x) = \int - F d x = \int \frac{3}{x + 1} d x = 3 ln (x + 1)$ J

Thus change in potential energy $Δ U = U {∣ ∣ ∣}_{x = 3} - U {∣ ∣ ∣}_{x = 1}$
$∴$ $Δ U = 3 ln (3 + 1) - [3 ln (1 + 1)] = 3 ln 4 - 3 ln 2 = 2.079$ J

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The conservative force exerted on a given particle as a function of position x is found to be F(x)=−3x+1 for positive x values, where x is given in meters.From x=1m to x=3m, by how much does the particle's potential energy change?

The correct option is D 2.079JThe force exerted by weird F=−3x+1 NPotential energy U(x)=∫−Fdx=∫3x+1dx=3ln(x+1) JThus change in potential energy ΔU=U∣∣∣x=3−U∣∣∣x=1∴ ΔU=3ln(3+1)−[3ln(1+1)]=3ln4−3ln2=2.079 J

The conservative force exerted on a given particle as a function of position $x$ is found to be $F (x) = \frac{- 3}{x + 1}$ for positive $x$ values, where $x$ is given in meters.
From $x = 1 m$ to $x = 3 m$ , by how much does the particle's potential energy change?