Question:
A constant horizontal force of 100 N is exerted on a 50 kg box initially at rest. When the box has moved a distance of 2 m, its speed is 2 m/s. The coefficient of kinetic friction between box and floor is (g = 10 m/s²)
A constant horizontal force of 100 N is exerted on a 50 kg box initially at rest. When the box has moved a distance of 2 m, its speed is 2 m/s. The coefficient of kinetic friction between box and floor is (g = 10 m/s²)
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Work-energy theorem is fastest for force + displacement + velocity problems.
Updated On: Jun 20, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the physical situation.
A block is pulled by a constant force while experiencing kinetic friction. The motion is along a horizontal surface, so net force depends on applied force minus friction force. We are given displacement and final speed, so energy or kinematics approach can be used.
Step 2: Identifying known quantities.
Mass \(m = 50 \, kg\), force \(F = 100 \, N\), displacement \(s = 2 \, m\), final velocity \(v = 2 \, m/s\), initial velocity \(u = 0\), and \(g = 10 \, m/s^2\). These values will help determine friction coefficient.
Step 3: Using work-energy theorem.
Work done by net force equals change in kinetic energy. This is the most direct method when force and displacement are given. So, \(W_{net} = \Delta KE\).
Step 4: Writing work done expression.
Net work = work by applied force − work by friction. Applied work = \(100 \times 2 = 200 \, J\). Friction force = \(\mu mg = \mu \times 50 \times 10 = 500\mu\). So friction work = \(500\mu \times 2 = 1000\mu\).
Step 5: Change in kinetic energy.
Final kinetic energy = \( \frac{1}{2}mv^2 = \frac{1}{2} \times 50 \times 4 = 100 \, J \). Initial KE = 0 since the box starts from rest.
Step 6: Forming equation.
So, \(200 - 1000\mu = 100\). Rearranging gives \(200 - 100 = 1000\mu\).
Step 7: Solving for coefficient of friction.
\(100 = 1000\mu \Rightarrow \mu = 0.1\).
Step 8: Final conclusion.
Thus, the coefficient of kinetic friction is 0.1.
A block is pulled by a constant force while experiencing kinetic friction. The motion is along a horizontal surface, so net force depends on applied force minus friction force. We are given displacement and final speed, so energy or kinematics approach can be used.
Step 2: Identifying known quantities.
Mass \(m = 50 \, kg\), force \(F = 100 \, N\), displacement \(s = 2 \, m\), final velocity \(v = 2 \, m/s\), initial velocity \(u = 0\), and \(g = 10 \, m/s^2\). These values will help determine friction coefficient.
Step 3: Using work-energy theorem.
Work done by net force equals change in kinetic energy. This is the most direct method when force and displacement are given. So, \(W_{net} = \Delta KE\).
Step 4: Writing work done expression.
Net work = work by applied force − work by friction. Applied work = \(100 \times 2 = 200 \, J\). Friction force = \(\mu mg = \mu \times 50 \times 10 = 500\mu\). So friction work = \(500\mu \times 2 = 1000\mu\).
Step 5: Change in kinetic energy.
Final kinetic energy = \( \frac{1}{2}mv^2 = \frac{1}{2} \times 50 \times 4 = 100 \, J \). Initial KE = 0 since the box starts from rest.
Step 6: Forming equation.
So, \(200 - 1000\mu = 100\). Rearranging gives \(200 - 100 = 1000\mu\).
Step 7: Solving for coefficient of friction.
\(100 = 1000\mu \Rightarrow \mu = 0.1\).
Step 8: Final conclusion.
Thus, the coefficient of kinetic friction is 0.1.
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