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Indian Navy SSR (Sailor) Work and Power — practice questions

Explore 17 free Work and Power questions in Indian Navy SSR (Sailor) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Work and Power questions and exercises with answers

Showing questions 1–17 of 17.

Kinetic friction f_k is given by:Read the answer and worked solution →A 5 kg block is pushed with 30 N along a frictionless surface. Acceleration:Read the answer and worked solution →A cricket ball of mass 150 g moving at (3i+4j) m/s is hit to -(3i+4j) m/s. The magnitude of momentum change isRead the answer and worked solution →As per your syllabus, a body moving on it must do which?Read the answer and worked solution →As per your syllabus, the maximum STATIC friction force on a block on a horizontal surface obeys which inequality?Read the answer and worked solution →True or false: what Aristotle got wrongRead the answer and worked solution →Complete Newton's first lawRead the answer and worked solution →Which law is this?Read the answer and worked solution →MomentumRead the answer and worked solution →True or false: inertia and forceRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Tap every statement the first law licensesRead the answer and worked solution →Flip and pair: name with contributionRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Order the historical argumentRead the answer and worked solution →

Worked practice questions

Kinetic friction f_k is given by:

  1. f_k = μ_k N
  2. f_k = μ_s N
  3. f_k = μ_k m
  4. f_k = 0
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Correct answer: A. f_k = μ_k N

Kinetic friction = μ_k × N, where μ_k is coefficient of kinetic friction and N is normal reaction. Unlike static friction, this is a definite value (when surfaces slide).

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A 5 kg block is pushed with 30 N along a frictionless surface. Acceleration:

  1. 0.17 m/s²
  2. 2 m/s²
  3. 6 m/s²
  4. 150 m/s²
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Correct answer: C. 6 m/s²

We have a = F/m. That is a = 30/5. So a = 6 m/s². On frictionless surface, all applied force goes into acceleration.

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A cricket ball of mass 150 g moving at (3i+4j) m/s is hit to -(3i+4j) m/s. The magnitude of momentum change is

  1. 0
  2. 3.0 kg·m/s
  3. 1.5 kg·m/s
  4. 0.75 kg·m/s
Show answer and solution

Correct answer: C. 1.5 kg·m/s

We have Δp = m(v−u). That is Δp = 0.15·(−2)(3i+4j). So Δp = −0.3(3i+4j). |Δp| = 0.3·√(9+16) = 0.3·5 = 1.5 kg·m/s.

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