Question: A $$2$$ kg body is dropped from height of $$1$$ m onto a&nbsp;spring of spring constant $$800$$ N/m as shown in&nbsp;the figure. A frictional force&nbsp;equivalent to $$0.4$$ kgwt&nbsp;acts on the body. What will be the speed of the body just&nbsp;before striking the spring? (Take $$g= 10\ m/s^2$$) Solution: We know that the change of kinetic energy is equal to work done by the system. &nbsp;Change of Kinetic energy $$=\dfrac{1}{2}mv^2-0=\dfrac{1}{2}mv^2$$ Work done $$=(mg-f)h$$ , where $$f=0.4 kgwt =0. 4\times10=4N$$, friction force. Now, $$\dfrac{1}{2}mv^2=(mg-f)h$$ $$ \dfrac{1}{2}2v^2=(20-4)1$$ $$v^2=16$$ $$v=4 m/s$$

Example:&nbsp; 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 $$\dfrac{1}{2}mv^2=f_r x+\dfrac{1}{2}kx^2$$$$\implies 5000x^2+15x-16=0$$$$\implies x=0.055m=5.5cm$$

Example: A $$500$$ gm ball moving at $$15$$ m/s slows down&nbsp;uniformly until it stops. If the ball travels $$15$$ m,&nbsp;what was the average net force applied while it&nbsp;was coming to a stop? Solution: Initial kinetic energy, $$KE_1=\dfrac{1}{2}mu^2=\dfrac{1}{2} \times 0.5 \times 15^2=56.25 J$$ Final kinetic energy, $$KE_2=\dfrac{1}{2}mv^2=0 J$$ Change in kinetic energy, $$\Delta KE=-56.25 J$$ By work energy theorem, work done equals change in kinetic energy. $$W= \Delta KE = -56.25 J$$ For a constant force, work done is $$W=F.s$$ $$F \times 15 = -56.25$$ $$F=-3.75N$$

Example: A particle of mass $$100$$ g is thrown vertically upwards with a speed of $$5$$ m/s. Find the work done by the&nbsp;force of gravity during the time the particle goes up. Solution: Change in kinetic energy is $$\Delta KE = 0 - \dfrac{1}{2} \times 0.1 \times 25 = -1.25 J$$ Using work-energy theorem, work done is given by, $$W=-1.25 J$$

Example: A body of mass $$2$$ kg is connected to a&nbsp;spring. Under influence of a constant&nbsp;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 = $$1\ &nbsp;N/m$$. Solution: Spring force on the body is given by, $$F=-kx$$ Work done by spring force is given by, $$W=\int F dx = \dfrac{1}{2} k (x_i^2-x_f^2)$$ $$W=\dfrac{1}{2} \times 1 (0.5^2-x_f^2)$$ By work-energy theorem, this equals the change in kinetic energy.&nbsp; $$\Delta KE = W$$ $$KE_f - 0 = \dfrac{1}{2}(0.5^2-x_f^2)$$ For maximum speed, $$KE_f$$ is maximum, i.e. $$x_f=0$$ $$\dfrac{1}{2}mv^2=0.125$$ $$v=0.353\ &nbsp;m/s$$

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Work Done by a Variable Force - Problem L1

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Friction and Spring Force acting on a body

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Use of Work Energy Theorem in Body Under Frictional Force

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Application of work-energy theorem for a body under constant external force

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Application of work-energy theorem in problems involving gravity

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Application of work-energy theorem in problems involving springs

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Work Done by a Variable Force - Problem L1

Work Done by a Variable Force - Problem L2