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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A block of mass 1.6 kg is attached to a horizontal spring that has a force constant of $1.0 \times 10^3 N/m$. The spring is compressed 2.0 cm and is then released from rest. Calculate the speed of the block as it passes through the equilibrium position $x= 0$ if the surface is frictionless.

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Answer: 0.5 m/s
The block starts with $v_i = 0$ at $x_i = -2 cm$. We want to find the speed of the block $v_f$ at $x_f = 0$.
The Work done by the spring with $x_{max} = x_i = - 2 cm$ is $W = \large\frac{1}{2}$$kx_{max}^2$
$\Rightarrow W = \large\frac{1}{2}$$ (1.0 \times 10^3 N/m) (-2 cm)^2 = 0.2 J$
Using the work–kinetic energy theorem, $W = \Delta KE + \Delta PE$, with 0, we set the change in kinetic energy of the block equal to the work done on it by the spring:
$\Rightarrow W = \large\frac {1}{2}$$ mv_f^2 - \large\frac{1}{2}$$ mv_i^2$
Solving for $v_f = \sqrt { v_i^2 + \large\frac{2W}{m}}$$ =\sqrt{0+ \large\frac{2 \times 0.2 J}{1.6 kg}}$$ = 0.5 m/s$
answered Aug 25, 2014 by balaji.thirumalai

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