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.
The engine of a plane exerts a force of 2 × 10⁴ N in the direction of motion when the plane's velocity is 540 km/h. Find the power of the engine in kilowatt.
Correct answer: (C) 3000 kW
Explanation: Step 1: v = 540 × 5/18 = 150 m/sec. Step 2: Power = F v = 2 × 10⁴ × 150 = 3 × 10⁶ watt = 3000 kW.
A body moves with constant speed 5 m/sec under a force of 100 N acting in the direction of motion. Find the power P developed by the force.
Correct answer: (C) 500 W
Explanation: Step 1: P = F·v. Step 2: P = 100 × 5. Step 3: P = 500. ∴ P = 500 W
A machine does 600 J of work in 30 s. Find its power P.
Correct answer: (C) 20 W
Explanation: Step 1: P = W / t. Step 2: P = 600 / 30. Step 3: P = 20. ∴ P = 20 W
A train of mass 250 tons moves on a horizontal railway against resistances of 12 kg.wt per ton of its mass. If the power of its engine is 480 horse, find its maximum velocity in km/h.
Correct answer: (C) 43.2 km/h
Explanation: Step 1: Resistance R = 12 × 250 = 3000 kg.wt; power = 480 × 75 = 36000 kg.wt.m/sec. Step 2: At maximum velocity the engine force = R, so v = 36000 ÷ 3000 = 12 m/sec. Step 3: v = 12 × 18/5 = 43.2 km/h.
A crane lifts a load of mass 50 kg to a height of 10 m in 20 s. Taking g = 9.8 m/sec², find the power output.
Correct answer: (B) 245 W
Explanation: Step 1: Work done against gravity = m g h = 50 × 9.8 × 10 = 4900 J. Step 2: Power = W ÷ t = 4900 ÷ 20. Step 3: = 245. ∴ Power = 245 W
A load of mass 20 kg is raised vertically at a constant speed of 2 m/sec. Taking g = 9.8 m/sec², find the power needed.
Correct answer: (A) 392 W
Explanation: Step 1: At constant speed the lifting force balances the weight: F = m g = 20 × 9.8 = 196 N. Step 2: Power = F v = 196 × 2. Step 3: = 392. ∴ Power = 392 W
A person of mass 60 kg climbs the stairs of a tower of height 49 m in 2 minutes. Find his average power in watt. (g = 9.8 m/sec²)
Correct answer: (C) 240.1 W
Explanation: Step 1: Work done against the weight = 60 × 9.8 × 49 = 28812 J. Step 2: Time = 2 × 60 = 120 sec. Step 3: Average power = 28812 ÷ 120 = 240.1 W.
A car of mass 1470 kg starts from rest and its engine works at a constant power of 90 horse. Neglecting resistances, find the time it takes to reach a velocity of 108 km/h. (1 horse = 735 watt)
Correct answer: (A) 10 sec
Explanation: Step 1: Work of the engine in time t = ∫₀ᵗ (90 × 735) dt = 66150 t joule. Step 2: v = 108 km/h = 30 m/sec, so the gain in kinetic energy = ½ × 1470 × 30² = 661500 J. Step 3: Work = change in kinetic energy: 66150 t = 661500, so t = 10 sec.
A car of mass 2 tons has an engine of power 96 horse and its maximum velocity on a horizontal road is 72 km/h. Find its maximum velocity up a road inclined to the horizontal at an angle whose sine is 1/50, if the resistance is the same on both roads.
Correct answer: (B) 64.8 km/h
Explanation: Step 1: Power = 96 × 75 = 7200 kg.wt.m/sec. On the horizontal road v = 20 m/sec, so R = 7200 ÷ 20 = 360 kg.wt. Step 2: Weight component down the slope = 2000 × 1/50 = 40 kg.wt. Step 3: Up the slope: F = 360 + 40 = 400 kg.wt, so v = 7200 ÷ 400 = 18 m/sec. Step 4: v = 18 × 18/5 = 64.8 km/h.
The power of an engine at time t seconds is (6t² + 2t) watt. Find the work done by the engine during the third second.
Correct answer: (D) 43 J
Explanation: Step 1: W = ∫ (power) dt. The third second is the interval from t = 2 to t = 3. Step 2: W = ∫₂³ (6t² + 2t) dt = [2t³ + t²] from 2 to 3. Step 3: = (54 + 9) − (16 + 4) = 63 − 20 = 43 J.
The first 10 questions in every subject are free, each with an explanation.