 # The rope and pulley have negligible mass, and the pulley is frictionle Carole Yarbrough 2021-12-16 Answered

The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the
8.00-kg block and the tabletop is $\mu k=0.250$. The blocks are released from rest. Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.

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Given:
$d=1.5m$
${\mu }_{k}=0.25$
${m}_{1}=8.0kg$
${m}_{2}=6.0kg$
Required: v
For horizontal mass
According to work-energy theorem the velocity is given by
${W}_{\ne t}=\frac{1}{2}{m}_{1}{v}^{2}⇒\left(1\right)$
But the net work is also given by
${W}_{\ne t}={W}_{t}-{W}_{f}={W}_{t}-{\mu }_{k}mgd⇒\left(2\right)$
Where : ${W}_{t}$ is the work due to tension.
From (1) and (2) get that
${W}_{t}-{\mu }_{k}{m}_{1}gd=\frac{1}{2}{m}_{1}{v}^{2}⇒\left(3\right)$
For vertical mass
According to work-energy theorem the velocity is given by
${W}_{\ne t}=\frac{1}{2}{m}_{2}{v}^{2}⇒\left(4\right)$
But the net work is also given by
${W}_{\ne t}={W}_{w}-{W}_{t}={m}_{2}gd-{W}_{t}⇒\left(5\right)$
From (4) and (5) get that
$-{W}_{t}+{m}_{2}gd=\frac{1}{2}{m}_{2}{v}^{2}⇒\left(6\right)$
Adding (3) and (6) to get
${m}_{2}gd-{\mu }_{k}{m}_{1}gd=\frac{1}{2}\left({m}_{1}+{m}_{2}\right){v}^{2}Rih>⇒\left(7\right)$
Substitution in (7) to get that
${v}^{2}=\frac{6×9.8×1.5-8×9.8×1.5×0.25}{7}⇒v=2.9\frac{m}{s}$

We have step-by-step solutions for your answer! Mollie Nash
Step1
Given that
mass ${m}_{1}=6$ kg
mass ${m}_{2}=8$ kg
coefficient of kinetic friction ${\mu }_{k}=0.250$
displacement of mass 6 kg y=1.50m
Step2
Assume that the uniform acceleration in both masses is "a" and tension in the rope is "T".
Now for mass ${m}_{1}$
effective downward force = 6g - T
${m}_{1}a=6g-T$
1) $6a=6g-T$
Now for mass ${m}_{2}$
friction force ( towards left) ${f}_{k}=\mid {\mu }_{k}{m}_{2}g$
${f}_{k}=\left(0.25\right)\left(8\right)\left(g\right)$
${f}_{k}=2g$
effective force on ${m}_{2}\text{( towards right)}=T-{f}_{k}$
${m}_{2}a=T-{f}_{k}$
2) $8a=T-2g$
Adding equation (1) and (2) as:
$\left(6a\right)+\left(8a\right)=\left(6g-T\right)+\left(T-2g\right)$
$14a=4g$
$a=\frac{4g}{14}$
$a=\frac{2g}{7}$
Step3
Effective force on ${m}_{1}$
${F}_{1}=6a$
${F}_{1}=\left(6\right)\left(\frac{2g}{7}\right)$
${F}_{1}=\frac{12g}{7}$
Work done on m, is
$W={F}_{1}y$
$W=\left(\frac{12g}{7}\right)\left(1.5\right)$
$W=\frac{18g}{7}$
Step4
Now by the use of work energy theorem
Work done = change in kinetic energy
$W=\frac{1}{2}m{v}^{2}$
where v = speed
$\frac{18g}{7}=\frac{1}{2}\left(6\right){v}^{2}$
${v}^{2}=\frac{36g}{42}$
g= gravitational acceleration ( $9.8\frac{m}{{s}^{2}}$)
$v=\sqrt{\frac{\left(36\right)\left(9.8\right)}{42}}$
$v=2.989\frac{m}{s}$

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