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.
A force of magnitude 50 N makes an angle of 60° with the direction of motion and moves the body a distance of 8 m. Calculate the work done.
Correct answer: (A) 200 J
Explanation: Step 1: W = F·d·cosθ. Step 2: cos60° = 0.5. Step 3: W = 50 × 8 × 0.5 = 200. ∴ W = 200 J
A force of magnitude 20 N acts on a body in the direction of its motion, moving it a distance of 5 m. Find the work done.
Correct answer: (A) 100 J
Explanation: Step 1: W = F·d·cosθ; the force is along the motion, so θ = 0° and cos0° = 1. Step 2: W = 20 × 5 × 1 = 100. ∴ W = 100 J
A body moves in a straight line while a resistance of magnitude 250 N acts on it. Find the work done by the resistance during a displacement of 0.4 km.
Correct answer: (A) −1 × 10⁵ J
Explanation: Step 1: The resistance acts opposite to the displacement, so its work is negative. Step 2: s = 0.4 km = 400 m. Step 3: W = −F s = −250 × 400 = −1 × 10⁵ J.
A horizontal force of 15 N pushes a box a distance of 6 m along a horizontal line in the direction of the force. Find the work done.
Correct answer: (B) 90 J
Explanation: Step 1: W = F·d·cosθ, with θ = 0° so cos0° = 1. Step 2: W = 15 × 6 × 1. Step 3: W = 90. ∴ W = 90 J
The force F = k i + 3j moves a particle in a straight line from A(1, 2) to B(5, −2). If the work done by F is zero, find k.
Correct answer: (A) 3
Explanation: Step 1: S = AB = B − A = 4i − 4j. Step 2: W = F · S = 4k − 12. Step 3: W = 0 gives 4k = 12, so k = 3 (F is perpendicular to the displacement).
Find, in joule, the work needed to lift 3 cubic metres of water to a height of 8 m (the mass of 1 cubic metre of water is 1000 kg, g = 9.8 m/sec²).
Correct answer: (C) 2.352 × 10⁵ J
Explanation: Step 1: Mass of the water = 3 × 1000 = 3000 kg, so its weight = 3000 × 9.8 = 29400 N. Step 2: The force needed (at least) equals the weight, acting upwards through 8 m. Step 3: W = 29400 × 8 = 235200 = 2.352 × 10⁵ J.
A constant force F = 4î + 3ĵ, measured in dyne, acts on a particle whose position vector at time t is r(t) = (t² + 1)î + (2t − 3)ĵ, measured in cm. Find the work done by F from t = 1 to t = 3, in joule.
Correct answer: (D) 4.4 × 10⁻⁶ joule
Explanation: S = r(3) − r(1) = (10î + 3ĵ) − (2î − ĵ) = 8î + 4ĵ cm. W = F · S = 32 + 12 = 44 dyne.cm = 44 erg. 1 joule = 10⁷ erg ⇒ W = 44 × 10⁻⁷ = 4.4 × 10⁻⁶ joule. (4.9 × 10⁻⁶ uses r(3) instead of the displacement.)
A worker of mass 65 kg carries some bricks and climbs a ladder whose top is 10 m above the ground. If the work done until he reaches the top is 8330 J, find the mass of the bricks. (g = 9.8 m/sec²)
Correct answer: (C) 20 kg
Explanation: Step 1: Work = (total weight) × height, so total weight × 10 = 8330 and total weight = 833 N. Step 2: Total mass = 833 ÷ 9.8 = 85 kg. Step 3: Mass of the bricks = 85 − 65 = 20 kg.
A body of mass 1 kg moves so that its displacement vector is S = 2t² i + (t² + 3t) j, where S is in metres and t in seconds. Find the work done by the moving force during the first 3 seconds of motion.
Correct answer: (B) 108 J
Explanation: Step 1: v = dS/dt = 4t i + (2t + 3) j and a = dv/dt = 4i + 2j (constant). Step 2: F = m a = 1 × (4i + 2j) = 4i + 2j newton. Step 3: Displacement in the first 3 s = S(3) − S(0) = 18i + 18j. Step 4: W = F · S = 4 × 18 + 2 × 18 = 108 J.
A horizontal force acts on a body of mass 20 kg at rest on a smooth horizontal plane and gives it a uniform acceleration of 0.98 m/sec² in its direction. If the work done by the force is 60 kg.wt.m, find the distance moved by the body.
Correct answer: (A) 30 m
Explanation: Step 1: F = m a = 20 × 0.98 = 19.6 N = 19.6 ÷ 9.8 = 2 kg.wt. Step 2: W = F × s, so 60 = 2 × s (both in kg.wt units). Step 3: s = 30 m.
The first 10 questions in every subject are free, each with an explanation.