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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 if a constant friction force of 4 N retards its motion from the moment it is released

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Answer: 039 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 $x_f = 0$.
We first calculate the kinetic energy lost because of friction and add this to the kinetic energy we calculated in the absence of friction.
Kinetic energy lost due to friction $\Delta K = -f_k d = -4 \times 2 \times 10^{-2} = -0.08 J$
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$
$\Rightarrow$, the final kinetic energy factoring in friction $K_f = 0.2 - 0.08 = 0.12J$
Now, $K_f = \large\frac{1}{2}$$mv_f^2$
$\Rightarrow v_f = \sqrt {\large\frac{2K_f}{m}}$$ = \sqrt{\large\frac{2 \times 0.12}{1.6}}$$ = 0.39 m/s$
answered Aug 25, 2014 by balaji.thirumalai
edited Aug 25, 2014 by balaji.thirumalai

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