Question: A 2 kg body is dropped from height of 1 m onto a spring of spring constant 800 N/m as shown in the figure. A frictional force equivalent to 0.4 kgwt acts on the body. What will be the speed of the body just before striking the spring? (Take g=10m/s2)
Solution: We know that the change of kinetic energy is equal to work done by the system. Change of Kinetic energy =21mv2−0=21mv2 Work done =(mg−f)h , where f=0.4kgwt=0.4×10=4N, friction force. Now, 21mv2=(mg−f)h 212v2=(20−4)1 v2=16 v=4m/s
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Use of Work Energy Theorem in Body Under Frictional Force
Example: A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10,000 N/m. What is the spring compression?
Solution: The total kinetic energy possessed by the block goes into the potential energy of the spring and the work done against friction.Let x be the compression of the spring.Thus 21mv2=frx+21kx2⟹5000x2+15x−16=0⟹x=0.055m=5.5cm
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Application of work-energy theorem for a body under constant external force
Example: A 500 gm ball moving at 15 m/s slows down uniformly until it stops. If the ball travels 15 m, what was the average net force applied while it was coming to a stop? Solution: Initial kinetic energy, KE1=21mu2=21×0.5×152=56.25J Final kinetic energy, KE2=21mv2=0J Change in kinetic energy, ΔKE=−56.25J By work energy theorem, work done equals change in kinetic energy. W=ΔKE=−56.25J For a constant force, work done is W=F.s F×15=−56.25 F=−3.75N
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Application of work-energy theorem in problems involving gravity
Example: A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. Find the work done by the force of gravity during the time the particle goes up. Solution: Change in kinetic energy is ΔKE=0−21×0.1×25=−1.25J Using work-energy theorem, work done is given by, W=−1.25J
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Application of work-energy theorem in problems involving springs
Example: A body of mass 2 kg is connected to a spring. Under influence of a constant external force, maximum elongation in the spring is 0.5 m. Now the external force is removed, find the maximum speed of the mass. Given spring constant = 1N/m. Solution: Spring force on the body is given by, F=−kx Work done by spring force is given by, W=∫Fdx=21k(xi2−xf2) W=21×1(0.52−xf2) By work-energy theorem, this equals the change in kinetic energy. ΔKE=W KEf−0=21(0.52−xf2) For maximum speed, KEf is maximum, i.e. xf=0 21mv2=0.125 v=0.353m/s