HomePhysics Class 11WORK ENERGY AND POWER MCQ TEST 2 WORK ENERGY AND POWER MCQ TEST 2 Work Engergy and Power 11 CBSE - MCQ test 1 Question 1. A block of mass m is placed on elevator moving down with accelertion g/3.What would be value of workdone by normal reaction acting on the block if elevator moves distance h downwards . mgh/3 -2mgh/3 2mgh/3 None of these Explanation Answer : -2mgh/3 Question 2. Force F acts on the free end of string as shwon pulley system below,if weight W moves up slowly a distance h.What would be value of work done by force F? Fh 2Fh Fh/2 2Fh/g Explanation Answer: 2Fh When W moves a distance h above and string at free end will move 2h below along the direction of force. Work done=F*2h=2Fh Question 3. Below figure is showing position time graph of mass 2Kg. What would be workdone by force for movement of mass from t=0 to t=4 sec. 8j 0 4j Can't detemine Explanation Answer:0 Dispalcement after t=4sec is 0. Question 4. A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly oppose to the motion.How much work does the cycle do on the road? 0J 2000J -2000J None Explanation Answer : 0 Question 5. Three forces of magnitude 14N,7N and 7N acting on a body in direction of vector ( 6i+2j+3k),(3i-2j+6k) and (2i-3j-6k)respevtively. The foreces remains constant while object moves from point A (2,-1,-3) to A (5,-1,1).What would be workdone on object. 150J 125J 0 75J Explanation Answer:75J Question:6 An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.20 m/s. What would be the minimum horsepower of the motor to be used ? 1.3hp 1.4hp 1.5 hp None Explanation Answer: 1.3 hp P = F v = mgv = (500 kg) (9.8 m/s2) (0.2 m/s) = 980 W Assuming no loss against friction etc., in the motor, the minimum horsepower of the motor is P = 980 W = 980/746 hp = 1.3 hp. Question:7 A body dropped from a height H reaches the ground with a speed of 1.2 √gH. Calculate the work done by air-friction. a force acts on it it moves through a certain distance it experiences an increase in energy through a mechanical influence. None of these Explanation Answer: It experiences an increase in energy through a mechanical influence. Question:8 A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with the horizontal. The friction coefficient between the block and the surface is µ. If the block travels at a uniform velocity, find the work done by this applied force during a displacement d of the block. μmg(dcos θ) /(cosθ - μsinθ) μmgd /(cosθ + μsinθ) μmg(dcos θ) /(cosθ + μsinθ) None Explanationn Answer: μmg(dcos θ) /(cosθ + μsinθ) Question:9 Static friction is variable. True False Expanation Answer: True Value of static friction changes when the value of applied force changes. Question:10 A man of 60 kg weight is standing at rest on a platform. He jumps up vertically a distance of 1 m and the platform at the same instant moves horizontally forward with the result that the man lands 1 meter behind the point on the platform where they took the jump, the total work done by the man at the instant he lands is .... 0 300 J 600 J None Explanation Answer:600 J Given, Mass of man = 60 Kg Distance covered by him as he jumps up = 1 m The formula of work done is Work = force x displacement W = F x d Now Force "F" = mg W = mg x d W = 60 x 10 x 1 W = 600 J Thus the work done is 600 J Question:11 A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table? 7.2J 10J 3.6J None Explanation Answer:3.6J Mass of the chain lying freely from the table=(Mass of chain*hanging length of chain)/Total length of chain =(4*0.60)/2=1.2kg The distance of center of mass of chain from the table= 0.5×0.6m=0.3m Thus the work done in pulling the chain=mgh=1.2×10×0.3J=3.6J Question:12 A body of mass 20 kg is initially at a height of 3 m above the ground. It is lifted to a height of 2 m from that position. Its increase in potential energy is: 120 J 90 J 392 J None of these Explanation Answer:392 J Given,Mass, m = 20kg Height m, H = 3m Potential Energy, (PE1) = mgH ,PE1 = 20 x 9.8 x 3 PE1 = 588 joules Now Height (h) is (3m + 2m)=5m (PE2) = mgh (PE2) = 20 x 9.8 x 5 (PE2) = 980 Joules Increase in Potential Energy = PE2 - PE1 = 980 - 588 = 392 Joules Question:13 The kinetic energy acquired by a mass m in travelling distance d, the starting from the rest under the action of a constant force, is directly proportional to: m 1/m m² Does not depend on mass Explanation Answer:100J Work done by external forces = Change in kinetic energy Since particle starts from rest, its initial Kinetic Energy is 0. Therefore, K.E.=Fd It is given that force is constant. We see that Kinetic energy is independent of mass. Question:14 A body of mass 10 kg moving at a height of 2 m, with uniform speed of 2 m/s. Its total energy is: 116 J 316 J 216 J None of these Explanation Answer:216 J Total energy of a body(E) = Kinetic Energy (KE) + Potential Energy (PE) Kinetic Energy (KE): The kinetic energy of an object can be calculated using the formula: KE = (1/2) * m * v^2 KE = (1/2) * 10 kg * (2 m/s)^2 KE = 20 J (Joules) Potential Energy (PE): The potential energy of an object near the surface of the Earth can be calculated using the formula: PE = m * g * h PE = 10 kg * 9.81 m/s^2 * 2 m PE = 196.2 J (Joules) Now, let's calculate the total energy: E = KE + PE E = 20 J + 196.2 J E = 216.2 J So, the total energy of the body is 216 Joules(APPROXIMATE). Question:15 Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is: 4:1 3:2 1:2 None Explanation Answer:1:2 We have, K.E=P²/2M P1²/2M = P2²/2M P1²/2*1 = P2²/2*4 P1²/P2²= 2/8 (P1/P2)² = 1/4 (P1/P2) = (1/2) Question:16 A body of mass 10 kg is initially at a height of 20 m above the ground. It falls to a height of 5 m above the ground. Its potential energy in the new position is: 490J 500J 1000J None of these Explanation Answer:490J PE = m * g * h PE = 10 kg * 9.8 m/s² * 5 m PE = 490 J (joules) So, the potential energy of the body in its new position, 5 m above the ground, is 490 joules. Question:17 A body of mass 100 kg falls from a height of 10 m. Its increase in kinetic energy is 1000 J 1800 J 9800 J None of these Explantion Answer:9800 J When an object is at some height the total energy in it is in the form of potential energy. When the body falls from a height below the potential energy gets converted into kinetic energy. Mass of body(m) = 100 kg Height(h)= 10m g=9.8m/s² Since kinetic energy is equal to loss of potential energy when it falls down, Increase in kinetic energy= loss in PE=100*9.8*10= 9800J Question:18 A gardener pushes a lawn roller through a distance of 20 m. If he applies a force of 20 kg wt with in a direction in blind at an angle 60 degree to the ground find the work done by the gardener. 4.36KJ 5.36KJ 3.36KJ 5.17KJ Explanation Answer:9800 J Force aplied, F = 20 kg-wt = 20×9.8N=196N Since,force is applied at an angle 60° with the ground. So, the horizontal component of the force that is responsible for displacement of the lawn roller= Fcos60°=196×0.5=98N So, work done=98×20=1960J Question:19 If force 𝐹⃗ = (𝑖̂+ 5𝑗̂+ 7𝑘̂) acts on a particle and displaces it through 𝑑⃗ = (6𝑖̂+ 9𝑘̂). Calculate the work done if the force is in newton and displacement is in metre 89 J 79 J 59 J 69 J Explanation Answer:69 J Force F = i + 5 j + 7 k Displacement s = 6 i + 9 k Work done = F . s = ( i + 5 j + 7 k ) . ( 6 i + 0 j + 9 k ) = 6 +0 +63 = 69 J. Question:20 A force F acting on an object varies with distance x as shown in figure below. The force is in newton and distance x is in metre. The work done by the force in moving the object from x = 0 to x = 6 m is: 12.5 J 13.5 J 0 None of these Explanation Answer:13.5 J The area under the force-displacement curve represents the work done on the object. Submit Newer Older