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Physics Question 31 – JEE-MAIN 2025

A block of mass 1 kg, moving along x with speed vi=10 m/s enters a rough region ranging from x=0.1 m to x=1.9 m. The retarding force acting on the block in this range is Fr=kx N, with k=10 N/m. Then the final speed of the block as it crosses rough region is.

The change in kinetic energy of the block is equal to the total work done by all forces acting on it.

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Ninja StrategyInitial Speed and Energy Loss Intuition

First, eliminate options that do not show a decrease in speed due to a retarding force. Then, use intuition about the magnitude of energy loss to rule out options that imply an unrealistically large or small change in kinetic energy.

Step 1: Calculate Work Done by Retarding Force✦ Active

The work done by the variable retarding force Fr=kx from x1=0.1 m to x2=1.9 m is given by the integral of force with respect to displacement. Given k=10 N/m.

W=x1x2Frdx=0.11.9(10x)dx W=10[x22]0.11.9=5[(1.9)2(0.1)2] W=5[3.610.01]=5×3.60=18 J
Step 2: Calculate Initial Kinetic Energy○ Expand

The initial kinetic energy of the block with mass m=1 kg and initial speed vi=10 m/s is:

KEi=12mvi2=12(1 kg)(10 m/s)2 KEi=12(1)(100)=50 J
Step 3: Apply Work-Energy Theorem to find Final Speed○ Expand

According to the Work-Energy Theorem, the work done on the block equals the change in its kinetic energy (W=KEfKEi). We can use this to find the final kinetic energy and then the final speed vf.

W=KEfKEi 18 J=12mvf250 J 12(1)vf2=5018=32 J vf2=64 m2/s2 vf=64=8 m/s
💡 Teacher's Secret Hint

Remember that speed is a scalar quantity and is always positive.

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