Physics: A uniform meter rule of mass 150g is pivoted at the 60cm mark. Calculate the mass which when hung at the 85cm mark will balance the ruler horizontally?. Please help

A uniform meter rule of mass 150g is pivoted at the

Ответ:

amypeace1978

09.01.2021

Physics

1) Spring constant = 2681.75 N/m

2)Velocity after launch = 13.33 m/s

Explanation:

Since the energy of system is conserved we shall equate initial energy and the energy at the top point of the ball

$E_{initial}=\frac{1}{2}kx^{2}\\\\\therefore E_{ini}=0.5\times k\times 0.27^{2}=0.03645k$

When the ball reaches it's maximum height it shall have only horizontal component of velocity while as the vertical component will become zero

Thus at the top most point the energy of the ball becomes

$E_{top}=\frac{1}{2}mv_{h}^{2}+mgh_{max}\\\\\therefore E_{top}=\frac{1}{2}m(v_{o}cos(\theta ))^{2}+mgh_{max}$

Applying values we get

$\therefore E_{top}=\frac{1}{2}\times 1.1\times (v_{o}cos(41^{o}))^{2}+1.1\times 9.81\times 3.9m$

Now since the motion of the ball is projectile motion the maximum height reached by the ball is given by

$h_{max}=\frac{v_{o}^{2}sin^{2}(\theta )}{2g}\\\\\therefore v_{o}=\sqrt{\frac{h_{max}\times 2g}{sin^{2}(\theta )}}\\\\\therefore v_{o}=\sqrt{\frac{3.9\times 2g}{sin^{2}(41)}}\\\\v_{o}=13.33m/s$

Using the velocity in the equation of energy at top we get

$E_{top}=97.75Joules$

Equating initial energy and the energy at the top we get

$0.03645k=97.75\\\\\therefore k=\frac{97.74}{0.03645}=2681.75N/m$

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A uniform meter rule of mass 150g is pivoted at the 60cm mark. Calculate the mass which when hung at the 85cm mark will balance the ruler horizontally?. Please help

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A uniform meter rule of mass 150g is pivoted at the 60cm mark. Calculate the mass which when hung at the 85￼cm mark will balance the ruler horizontally?. Please help

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A uniform meter rule of mass 150g is pivoted at the 60cm mark. Calculate the mass which when hung at the 85cm mark will balance the ruler horizontally?. Please help