A 2 kg block slides on a horizontal floor with a speed of 4 m/s . It strikes an uncompressed spring and compresses it till the block is motionless. Th - Physics MCQ AIEEE 2007 2026 | ExamTest.live

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A 2 kg block slides on a horizontal floor with a speed of 4 m/s . It strikes an uncompressed spring and compresses it till the block is motionless. The kinetic friction force is 15 N and the spring constant is 10, 000 N/m . The spring compresses by:

Solution & Step-by-step Explanation

By the Work-Energy Theorem: W_{n e t} = Δ K E .Work is done by the spring force and the friction force. W_{s p r in g} + W_{f r i c t i o n} = K E_{f ina l} - K E_{ini t ia l} - \frac{1}{2} k x^{2} - f_{k} x = 0 - \frac{1}{2} m v^{2} \frac{1}{2} (10000) x^{2} + 15 x = \frac{1}{2} (2) (4^{2}) 5000 x^{2} + 15 x - 16 = 0 Using the quadratic formula x = \frac{- b \pm b ^{2} - 4 a c}{2 a} : x = \frac{- 15 \pm 1 5 ^{2} - 4 ( 5000 ) ( - 16 )}{10000} x = \frac{- 15 \pm 225 + 320000}{10000} \approx \frac{- 15 + 565.88}{10000} x \approx \frac{550.88}{10000} = 0.055 m = 5.5 cm

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A 2 kg block slides on a horizontal floor with a speed of 4 m/s . It strikes an uncompressed spring and compresses it till the block is motionless. The kinetic friction force is 15 N and the spring constant is 10, 000 N/m . The spring compresses by:

5.5 cm

2.5 cm

11.0 cm

8.5 cm

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