Applied Mathematics — Second: Dynamics · Mathematics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 36 questions on this lesson, each with an explanation.
Find the kinetic energy of a body of mass 2 kg moving with speed 10 m/sec.
Correct answer: (C) 100 J
Explanation: Step 1: Kinetic energy = ½·m·v². Step 2: Kinetic energy = ½ × 2 × 10² = ½ × 2 × 100. Step 3: Kinetic energy = 100. ∴ Kinetic energy = 100 J
A body of mass 5 kg is at a height of 4 m above the ground. Taking g = 9.8 m/sec², find its potential energy.
Correct answer: (A) 196 J
Explanation: Step 1: Potential energy = m·g·h. Step 2: Potential energy = 5 × 9.8 × 4. Step 3: Potential energy = 196. ∴ Potential energy = 196 J
A body of mass 50 gm moves with velocity v = 30i + 40j, where the magnitude of the velocity is in cm/sec. Find its kinetic energy in joule.
Correct answer: (B) 6.25 × 10⁻³ J
Explanation: Step 1: ‖v‖ = √(30² + 40²) = 50 cm/sec. Step 2: T = ½ m ‖v‖² = ½ × 50 × 2500 = 62500 erg. Step 3: 1 joule = 10⁷ erg, so T = 62500 ÷ 10⁷ = 6.25 × 10⁻³ J.
A body of mass 2 kg starts from rest and reaches a speed of 6 m/sec. Using the work–energy theorem, find the net work done on the body.
Correct answer: (A) 36 J
Explanation: Step 1: Work–energy theorem: net work = change in kinetic energy = ½·m·v² − ½·m·u², with u = 0. Step 2: Net work = ½ × 2 × 6² = ½ × 2 × 36. Step 3: Net work = 36. ∴ Net work = 36 J
A bullet of mass 20 gm is fired with velocity 300 m/sec into a thick barrier and embeds in it to a depth of 15 cm. Find the resistance of the barrier, assumed constant.
Correct answer: (B) 6000 N
Explanation: Step 1: Initial kinetic energy = ½ × 0.02 × 300² = 900 J; final kinetic energy = 0. Step 2: Work–energy principle: 0 − 900 = −R × 0.15. Step 3: R = 900 ÷ 0.15 = 6000 N.
A body of mass 5 kg slows down from 10 m/sec to 4 m/sec under a resisting force. Find the magnitude of the work done by that resisting force.
Correct answer: (D) 210 J
Explanation: Step 1: Work–energy theorem: work of the force = change in kinetic energy = ½·m·v² − ½·m·u². Step 2: Change in kinetic energy = ½ × 5 × (4² − 10²) = ½ × 5 × (16 − 100) = ½ × 5 × (−84) = −210 J. Step 3: The negative sign shows the force opposes motion; its magnitude of work is 210. ∴ Work = 210 J
A car starts from rest and descends a slope of length 200 m whose top is 20 m above its bottom. If 5/8 of the potential energy lost is used in overcoming the resistances, find the speed of the car at the bottom of the slope. (g = 9.8 m/sec²)
Correct answer: (B) 7√3 m/sec
Explanation: Step 1: Lost potential energy = m × 9.8 × 20 = 196 m. Step 2: Lost potential energy = gain in kinetic energy + work against resistances, so the gain in kinetic energy = 3/8 × 196 m = 73.5 m. Step 3: ½ m v² = 73.5 m, so v² = 147. Step 4: v = √147 = 7√3 m/sec.
A body of mass 2 kg is released from rest and falls freely through 20 m. Taking g = 9.8 m/sec², use the work–energy principle to find its kinetic energy after it has fallen 20 m.
Correct answer: (D) 392 J
Explanation: Step 1: Work done by gravity = m g h = 2 × 9.8 × 20 = 392 J. Step 2: Work–energy principle: work done = change in kinetic energy = T − 0, and gravity is the only force doing work. Step 3: T = 392. ∴ Kinetic energy = 392 J
A bullet is fired into a target of thickness 12 cm and comes out of the other side with half the velocity it had when it entered. What is the least thickness of a target of the same material that would stop the bullet if it is fired with the same velocity? (Assume the resistance is constant.)
Correct answer: (B) 16 cm
Explanation: Step 1: Let the entry kinetic energy be T. It leaves with half the velocity, so with ¼ T: it loses ¾ T in 12 cm. Step 2: Work against the constant resistance R: R × 12 = ¾ T, so R = T/16. Step 3: To stop it: R × d = T, so d = 16 cm.
A body of mass 4 kg moving at 12 m/sec is brought to rest by a constant resisting force of 24 N. Find the distance it travels before stopping.
Correct answer: (D) 12 m
Explanation: Step 1: Work done against the body = change in kinetic energy = ½·m·u² = ½ × 4 × 12² = 288 J. Step 2: This work equals F·d, so d = 288 / 24. Step 3: d = 12. ∴ d = 12 m
The first 10 questions in every subject are free, each with an explanation.