Question:
A 1 kg box placed at the origin starts sliding along x-axis under the action of a force \(\vec{F} = m \vec{a}\). Its acceleration as a function of x is given by \(a(x) = \beta x\) where \(\beta = 5 \, \text{s}^{-2}\). Find the work done by \(\vec{F}\) in moving the box from \(x = 2 \, \text{cm}\) to \(x = 5 \, \text{cm}\).
A 1 kg box placed at the origin starts sliding along x-axis under the action of a force \(\vec{F} = m \vec{a}\). Its acceleration as a function of x is given by \(a(x) = \beta x\) where \(\beta = 5 \, \text{s}^{-2}\). Find the work done by \(\vec{F}\) in moving the box from \(x = 2 \, \text{cm}\) to \(x = 5 \, \text{cm}\).
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For variable acceleration along x-axis, work done by force can be calculated using \(W = \int F dx = m \int a(x) dx\).
Updated On: Jul 18, 2026
- \(52.5 \times 10^{-4} \, \text{J}\)
- \(105.5 \times 10^{-4} \, \text{J}\)
- \(17.0 \times 10^{-4} \, \text{J}\)
- \(34.0 \times 10^{-4} \, \text{J}\)
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The Correct Option is A
Solution and Explanation
Step 1: Relate work and force.
Work done by force is:
\[ W = \int_{x_1}^{x_2} F \, dx = \int_{x_1}^{x_2} m a(x) \, dx \]
Here \(m = 1 \, \text{kg}\), \(a(x) = \beta x\), \(x_1 = 0.02 \, \text{m}\), \(x_2 = 0.05 \, \text{m}\).
Step 2: Substitute values.
\[ W = \int_{0.02}^{0.05} 1 \cdot 5 x \, dx = 5 \int_{0.02}^{0.05} x \, dx \]
Step 3: Integrate.
\[ \int x \, dx = \frac{x^2}{2} \implies W = 5 \left[ \frac{x^2}{2} \right]_{0.02}^{0.05} = \frac{5}{2} (0.05^2 - 0.02^2) \]
Step 4: Calculate the result.
\[ 0.05^2 = 0.0025, \quad 0.02^2 = 0.0004 \]
\[ 0.0025 - 0.0004 = 0.0021 \]
\[ W = \frac{5}{2} \cdot 0.0021 = 0.00525 \, \text{J} = 52.5 \times 10^{-4} \, \text{J} \]
Step 5: Verify units.
Force in N, displacement in m, work in J. Units are consistent.
Step 6: Final conclusion.
Hence, the work done by the force is:
\[ \boxed{52.5 \times 10^{-4} \, \text{J}} \]
Work done by force is:
\[ W = \int_{x_1}^{x_2} F \, dx = \int_{x_1}^{x_2} m a(x) \, dx \]
Here \(m = 1 \, \text{kg}\), \(a(x) = \beta x\), \(x_1 = 0.02 \, \text{m}\), \(x_2 = 0.05 \, \text{m}\).
Step 2: Substitute values.
\[ W = \int_{0.02}^{0.05} 1 \cdot 5 x \, dx = 5 \int_{0.02}^{0.05} x \, dx \]
Step 3: Integrate.
\[ \int x \, dx = \frac{x^2}{2} \implies W = 5 \left[ \frac{x^2}{2} \right]_{0.02}^{0.05} = \frac{5}{2} (0.05^2 - 0.02^2) \]
Step 4: Calculate the result.
\[ 0.05^2 = 0.0025, \quad 0.02^2 = 0.0004 \]
\[ 0.0025 - 0.0004 = 0.0021 \]
\[ W = \frac{5}{2} \cdot 0.0021 = 0.00525 \, \text{J} = 52.5 \times 10^{-4} \, \text{J} \]
Step 5: Verify units.
Force in N, displacement in m, work in J. Units are consistent.
Step 6: Final conclusion.
Hence, the work done by the force is:
\[ \boxed{52.5 \times 10^{-4} \, \text{J}} \]
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