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Under the action of a force a 2kg body moves

WebUnder the action of a force a 2kg body moves such that its position x in meters as a function of time t is given by x = 4t4 + 3 . Then work done by the force in first two seconds is 1691 33 NTA Abhyas NTA Abhyas 2024 Work, Energy and Power Report Error A 6J B 10J C 7J D 64J Solution: x = 4t4 +3 dtdx = 44t3 + 0 v = t3 Using work energy theorem WebA heavy suitcase S of mass 50 kg is moving along a horizontal floor under the action of a force of magnitude P newtons. The force acts at 30° to the floor, as shown in the diagram above, and S moves in a straight line at constant speed. The suitcase is modelled as a particle and the floor as a rough horizontal plane.

Under the action of a force, a 2 kg body moves such that

Web24 Mar 2024 · A body of mass 2 k g initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s. (b) work done by friction in 10 s , (c) work done by the net force on the body in 10 s , (d) change in kinetic energy of the body in 10 s , Web25 Jul 2024 · Also Read : A particle of mass m0 ascends from the earth’s surface with zero velocity due to the action of two…. A body moves a distance of 10 m under the action of force F = 10 N. If the work done is 25 J, the…. If action force is gravitational force, then reaction force. A particle is displaced from a position (2i -j + k ) to another ... boney m en chile https://stebii.com

Under the action of a force, a `2 kg` body moves such that its …

WebQ: Under the action of force 2 kg body moves such that its position ‘x’ varies as a function of time t given by x= t 3 /3, x is the meter and t in second. Calculate the workdone by the … Web17 Aug 2024 · Best answer The acceleration produced in the body by the applied force is given by Newton’s second law of motion as: a' = F/m = 7/2 = 3.5 m/s2 Work done by the net force, Wnet= 5.04 ×126 = 635 J From the first equation of motion, final velocity can be calculated as: v = u + at = 0 + 2.52 × 10 = 25.2 m/s ← Prev Question Next Question → goblins clash of clans

A particle of mass `2 kg` moves in the `xy` plane under …

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Under the action of a force a 2kg body moves

A block of mass 2 kg initially at rest moves under the action of an ...

WebQ. Under action of force, a 2 k g body moves such that its position x as function of time t is given by x = α t 2 /2 where x is in meters, t is in seconds and α = 1 m / s 2. The work done by the force in the first two seconds is WebUnder the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x = t 3/3, where x is in meter and t in second. The work done by the force in the …

Under the action of a force a 2kg body moves

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WebSolutions for Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x = t3/3, where x is in meters and t is in sec. The work done by … WebAnswer (1 of 3): x = t^3/3 v = dx/dt = t^2 a = dv/dt = 2t F = ma = 4t dx = t^2 dt Work done by the force = INTEG [t = 0 to 2] {F dx } = INTEG [t = 0 to 2] {4t^3} = 16 J x = t^3/3 v = dx/dt = …

WebA block of mass 2 kg initially at rest moves under the action of an applied horizontal force of 6 N on a rough horizontal surface. The coefficient of friction between block and surface is 0.1. The work done by the applied force in 10 s is (Take g = 10 ms-2) Q. WebUnder the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x = t^33 , where x is in meter and t in second. The work done by the force in …

WebA body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, WebA particle P of mass 0.5 kg moves under the action of a single force F newtons. At time t seconds, the velocity v m s–1 of P is given by . v = 3t2i + (1 – 4t)j. Find (a) the acceleration of P at time t seconds, (2) (b) the magnitude of F when t = 2. (4) (Total 6 marks) 4. A particle P of mass 0.5 kg is moving under the action of a single ...

WebQuestion From – Cengage BM Sharma MECHANICS 1 WORK, POWER & ENERGY JEE Main, JEE Advanced, NEET, KVPY, AIIMS, CBSE, RBSE, UP, MP, BIHAR BOARDQUESTION TEXT: …

Web25 Mar 2024 · Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x = (t^ (3))/ (3) where x is in metre and t in second. The work … goblinscomic twitterWeb24 Dec 2024 · A body of mass 1 kg begins to move under the action of a time dependent force vector F = (2t i + 3t 2 j)N,where vector i and j are unit vectors along x and y axis. … boney menWebUnder the action of a force, a `2 kg` body moves such that its position x as a function of time is given by `x =(t^(3))/(3)` where x is in metre and t in sec...... goblins clash a ramaWeb26 Jan 2024 · A particle of mass `2 kg` moves in the `xy` plane under the action of a constant force `vec (F)` where `vec (F)=hat (i)-hat (j)`. Initially the velocity of the particle is … goblins churWebMechanics. work-energy-and-power. under the action of a force ,a 2 kg body moves such that its position x as a function of time is given by x=t³/3.where t is in seconds and x in metre .the work done by force in first 2 seconds is? i diffrentiated x=t³/3 twice to get the equation for acceleration and got it 4 m/s^2 then force=4*2=8 N now as W ... boney m farrellWeb3 Apr 2024 · Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by where x is in metre and t in second. The work done by the force in first two seconds is (1) 1600 (2) 160 (3) 160 (4) 16/9 Theory : work energy theorem: Solution : given: mass of the body = 2 kg position as a function of time is given by : boney m feat. maizie williamsWebUnder action of force, a 2kg body moves such that its position x as function of time t is given by x = αt2/2 where x is in meters, t is in seconds and α = 1m/s2. The work done by … boney m featuring maizie williams