Question

A 12-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 0.80 kN/m). The block is initially at rest at its equilibrium position when a force (magnitude P = 100 N) acting parallel to the surface is applied to the block, as shown. What is the speed of the block (in m/s) when it is 13 cm from its equilibrium position?

Answer 1

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The law of conservation of energy says that energy will be a constant quantity. The work done is given by the equation

$W= \dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2$--------------(1)

where,

m= mass of the particle

W= work done

K= constant

x= compression length

v= velocity

$W=P.x$----------------(2)

here P= force applied

by using (1) and (2) we get,

$P.x= \dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2$----------------------(3)

solve eq'' (3) for v we get,

$v= \sqrt{\dfrac{2P.x}{m}-\dfrac{kx^2}{m}}$

and now substitude all values P.x= 80N, K= 0.80 kN/m, x= 13cm, m= 12kg

after all calculaton we get the answer{ refer the attachment for calculation}

v=0.7788m/s

Answer 2

Given:

Mass of the block $=12kg$

Spring constant $(k)=0.80N/m$

External force $=100N$

To find:

Speed of the block.

Explanation:

• According to law of conservation of energy, energy in a system stays conserved always.

• For a spring, the force applied is stored as its potential energy which is given by    $E=\frac{1}{2}kx^{2}$ where, x is the elongation produced in the spring and $k$ is the spring constant of the spring.

Solution :

According to the question, we have been given that on a block of mass $12kg$ attached to the spring, a force is applied for 13 cm, we know, that the energy produced by the compression of the spring will be stored as the potential energy in the spring and the excess energy will be given out as the kinetic energy.

$E=\frac{1}{2}kx^{2}+\frac{1}{2}mv^{2}$

Where, $m$ is mass of the block ; and

$v$ is the velocity of the excess energy converted to kinetic energy.

We know that the energy is the capacity to do work and work done is the force applied on the body and the displacement it thus produces.

Hence, here,

$E=F$ × $x$

The equation becomes,

$Fx=\frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

We know,

$F=100N$    ;  $x=0.13m$     ;  $m=12kg$     ;  $k=0.8N/m$

Substituting he given values, we get

$100(0.13)=\frac{1}{2}(0.8)(0.13)^{2}+\frac{1}{2}(12)v^{2}$

$13=\frac{1}{2}(0.01352)+6v^{2}$

$13=0.00676+6v^{2}$

$12.99324=6v^{2}$

$v^{2}=2.16554$

$v=1.471577 m/s$   ≅  $1.48m/s$

Final answer:

Hence, the velocity of the block will be 1.48 m/s.