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MP DTE Work and Energy — practice questions — page 2

Explore 102 free Work and Energy questions in MP DTE Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Work and Energy questions and exercises with answers

Showing questions 31–60 of 102.

A 50 kg person stands on a scale in an elevator accelerating UP at 2 m/s². Scale reading:Read the answer and worked solution →Two equal masses (3 kg each) connected by a light string over a frictionless pulley. One on smooth horizontal table, other hangs over edge. Acceleration?Read the answer and worked solution →A pendulum of mass m swings to amplitude angle θ. At the lowest point of swing, the TENSION in the string equals:Read the answer and worked solution →A car of mass 1000 kg moving at 36 km/h is stopped by brakes in 5 seconds. Average force exerted by brakes:Read the answer and worked solution →Newton's first law of motion is also known as the law ofRead 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 →Conservation of linear momentum in a collision follows directly fromRead the answer and worked solution →A force F acts on a 2 kg body for 3 s, changing its speed from 2 m/s to 8 m/s along the line of F. The magnitude of F isRead the answer and worked solution →The maximum static friction force on a body is f = μN. Static friction is best described asRead the answer and worked solution →A 10 kg block on a frictionless surface is pulled by 20 N horizontally and 20 N at 60° above horizontal. The horizontal acceleration isRead the answer and worked solution →Impulse equalsRead the answer and worked solution →A body in equilibrium under several forces hasRead the answer and worked solution →A 0.5 kg ball hits a wall at 10 m/s and rebounds at 10 m/s. The impulse on the ball isRead the answer and worked solution →Newton's second law in vector form isRead the answer and worked solution →A lift accelerates upward at 2 m/s². The apparent weight of a 50 kg person (g=10) isRead the answer and worked solution →Two blocks of 2 kg and 3 kg connected by a string are pulled by 10 N on the 3 kg block (frictionless). The acceleration isRead the answer and worked solution →The SI unit of force, the newton, equalsRead the answer and worked solution →A passenger standing in a bus is thrown backward when the bus suddenly starts moving forward. This is best explained by:Read the answer and worked solution →One newton is defined as the force that gives a mass of 1 kg an acceleration of:Read the answer and worked solution →The linear momentum of a body is defined as the product of its:Read the answer and worked solution →The impulse of a force has the same dimensions as:Read the answer and worked solution →The action and reaction forces of Newton's third law:Read the answer and worked solution →For a given pair of surfaces, the coefficient of kinetic friction is usually:Read the answer and worked solution →A constant force of 20 N acts on a 5 kg body initially at rest. Its speed after 3 s is:Read the answer and worked solution →A 0.15 kg ball moving at 12 m s⁻¹ is hit straight back along its path with the same speed. The magnitude of the impulse on the ball is:Read the answer and worked solution →A 0.04 kg bullet moving at 90 m s⁻¹ is stopped after penetrating 0.6 m into a block. The average resistive force is:Read the answer and worked solution →A block just begins to slide down an incline when the angle of inclination is 30°. The coefficient of static friction between block and incline is:Read the answer and worked solution →A 0.02 kg bullet is fired with a speed of 300 m s⁻¹ from a 4 kg gun. The recoil speed of the gun is:Read the answer and worked solution →A 50 kg person stands on a weighing scale inside a lift accelerating upward at 2 m s⁻² (g = 10 m s⁻²). The scale reads:Read the answer and worked solution →A 0.5 kg stone tied to a string is whirled in a horizontal circle of radius 2 m at a speed of 4 m s⁻¹. The tension in the string is:Read the answer and worked solution →

Worked practice questions

A 50 kg person stands on a scale in an elevator accelerating UP at 2 m/s². Scale reading:

  1. 400 N
  2. 500 N
  3. 600 N
  4. 700 N
Show answer and solution

Correct answer: C. 600 N

Apparent weight = m(g + a) = 50 × (10 + 2) = 50 × 12 = 600 N. (Going up: scale reads MORE than at rest weight of 500 N. Going down/free fall: less.)

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Two equal masses (3 kg each) connected by a light string over a frictionless pulley. One on smooth horizontal table, other hangs over edge. Acceleration?

  1. g/2 = 5 m/s²
  2. g/4 = 2.5 m/s²
  3. g = 10 m/s²
  4. Zero
Show answer and solution

Correct answer: A. g/2 = 5 m/s²

**Hanging mass:** T = m(g - a). Mass on table: T = ma. **Equating:** m(g - a) = ma → g - a = a → a = g/2 = 5 m/s². (Atwood-like setup with no friction.)

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A pendulum of mass m swings to amplitude angle θ. At the lowest point of swing, the TENSION in the string equals:

  1. mg
  2. mg(2 − cos θ)
  3. mg(3 − 2cos θ)
  4. 2mg sin θ
Show answer and solution

Correct answer: C. mg(3 − 2cos θ)

**At the lowest point:** T − mg = mv²/L (centripetal). v² at bottom = 2gL(1 − cos θ) (energy conservation from amplitude). We have So T = mg + 2mg(1 − cos θ). So So T = mg(3 − 2cos θ). For θ = 60°: T = mg(3 − 1) = 2mg.

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