# The rope and pulley have negligible mass, and the pulley is frictionle

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.

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Bertha Jordan
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
${v}^{2}=\frac{6×9.8×1.5-8×9.8×1.5×0.25}{7}⇒v=2.9\frac{m}{s}$

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 ( $9.8\frac{m}{{s}^{2}}$)
$v=\sqrt{\frac{\left(36\right)\left(9.8\right)}{42}}$
$v=2.989\frac{m}{s}$