Question

A curve is given between the potential energy \(U\) of a particle and its position on the \(x\)-axis as shown. Given: \[ \tan\theta_1 = 1,\qquad \tan\theta_2 = 3,\qquad \tan\theta_3 = -\frac{1}{2} \] If \(F_{AB}\) is the force acting on the particle during motion from \(A\) to \(B\), similarly \(F_{BC}\), \(F_{CD}\) and \(F_{DE}\) are the forces during \(B\) to \(C\), \(C\) to \(D\) and \(D\) to \(E\) respectively, arrange the magnitudes of these forces in decreasing order.

• A. \(F_{BC}>F_{AB}>F_{CD}>F_{DE}\)
• B. \(F_{BC}>F_{AB}>F_{DE}>F_{CD}\)
• C. \(F_{AB}>F_{BC}>F_{DE}>F_{CD}\)
• D. \(F_{BC}>F_{DE}>F_{AB}>F_{CD}\)

Hint:

For potential energy graphs:
Force is given by the negative slope of the \(U\)--\(x\) graph
Steeper slope means larger force
Flat regions correspond to zero force

Answer 1

The Correct Option is B

To solve this problem, we need to understand the relationship between potential energy \(U\) and force \(F\). The force acting on a particle is given by the negative gradient of the potential energy with respect to position:

\(F = -\frac{dU}{dx}\) 

This equation tells us the force is related to the slope of the potential energy curve. The greater the slope, the greater the force.

The problem gives us the slopes of segments \(AB\), \(BC\), \(CD\), and \(DE\) as the tangents of the angles. Therefore, the magnitude of the force is proportional to these values:

• \(\tan\theta_1 = 1\) for \(AB\)
• \(\tan\theta_2 = 3\) for \(BC\)
• \(\tan\theta_3 = -\frac{1}{2}\) for \(DE\)
• Slope for \(CD\) is \(0\) because it's constant

The force magnitudes, in terms of these slopes, are therefore:

• \(|F_{AB}| = 1\)
• \(|F_{BC}| = 3\)
• \(|F_{DE}| = \frac{1}{2}\)
• \(|F_{CD}| = 0\)

Arranging these in decreasing order gives:

• \(|F_{BC}| > |F_{AB}| > |F_{DE}| > |F_{CD}|\)

Thus, the correct answer is \(F_{BC}>F_{AB}>F_{DE}>F_{CD}\).