Question Number 25253 by NECx last updated on 07/Dec/17
$${A}\:\mathrm{100}{kg}\:{man}\:{lowers}\:{himself}\:{to} \\ $$$${the}\:{ground}\:{from}\:{a}\:{height}\:{of}\:\mathrm{10}{m} \\ $$$${by}\:{means}\:{of}\:{a}\:{rope}\:{passed}\:{over}\:{a} \\ $$$${frictionless}\:{pulley}\:{and}\:{the}\:{other} \\ $$$${end}\:{attached}\:{to}\:{a}\:\mathrm{70}{kg}\:{sandbag}. \\ $$$$\left.{a}\right){with}\:{what}\:{speed}\:{does}\:{the}\:{man}\: \\ $$$${hit}\:{the}\:{ground}? \\ $$$$\left.{b}\right){could}\:{he}\:{have}\:{done}\:{anything}\:{to} \\ $$$${reduce}\:{the}\:{speed}? \\ $$
Commented by NECx last updated on 07/Dec/17
$${please}\:{do}\:{well}\:{to}\:{interprete}\:{with}\:{a} \\ $$$${diagram}.{That}\:{was}\:\mathrm{1}\:{of}\:{my}\: \\ $$$${challenge}\:{while}\:{I}\:{tried}\:{solving}\:{it}. \\ $$$${Thanks} \\ $$
Answered by mrW1 last updated on 07/Dec/17
$$\left.{a}\right) \\ $$$${acceleration}\:{of}\:{man}\:{and}\:{sandbag}\:{is} \\ $$$${a}=\frac{\left({m}_{{M}} −{m}_{{S}} \right){g}}{{m}_{{M}} +{m}_{{S}} }=\frac{\mathrm{100}−\mathrm{70}}{\mathrm{100}+\mathrm{70}}×\mathrm{10}=\frac{\mathrm{30}}{\mathrm{17}}\:{m}/{s}^{\mathrm{2}} \\ $$$$ \\ $$$${speed}\:{of}\:{them}\:{is} \\ $$$${v}=\sqrt{\mathrm{2}{ah}}=\sqrt{\mathrm{2}×\frac{\mathrm{30}}{\mathrm{17}}×\mathrm{10}}=\mathrm{5}.\mathrm{96}\:{m}/{s} \\ $$$$\left.{b}\right) \\ $$$${the}\:{man}\:{can}\:{reduce}\:{his}\:{speed}\:{if}\:{he}\:{also} \\ $$$${holds}\:{the}\:{rope}\:{on}\:{the}\:{side}\:{of}\:{sandbay} \\ $$$${and}\:{releases}\:{this}\:{rope}\:{gradually}. \\ $$
Commented by NECx last updated on 07/Dec/17
$${how}\:{did}\:{you}\:{derive}\:{the}\:{formula} \\ $$$${for}\:{the}\:{acceleration}? \\ $$
Commented by mrW1 last updated on 07/Dec/17
$${man}\:{and}\:{sandbag}\:{move}\:{together} \\ $$$${as}\:{a}\:{single}\:{object}.\:{total}\:{mass}\:{of}\:{them} \\ $$$${is}\:{M}={m}_{{Man}} +{m}_{{Sand}} .\:{The}\:{total}\:{force}\:{which} \\ $$$${acts}\:{on}\:{them}\:{is}\:{F}={m}_{{Man}} {g}−{m}_{{Sand}} {g}=\left({m}_{{Man}} −{m}_{{Sand}} \right){g}, \\ $$$${therefore}\:{the}\:{acceleration}\:{of}\:{them} \\ $$$${is}\:{a}=\frac{{F}}{{M}}=\frac{\left({m}_{{M}} −{m}_{{S}} \right){g}}{{m}_{{M}} +{m}_{{S}} }. \\ $$$$ \\ $$$${An}\:{other}\:{way}\:{is}: \\ $$$${let}\:{T}={tension}\:{in}\:{rope} \\ $$$${m}_{{Man}} {a}={m}_{{Man}} {g}−{T}\:\:\:\:...\left({i}\right) \\ $$$${m}_{{Sand}} {a}={T}−{m}_{{Sand}} {g}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\left({m}_{{M}} +{m}_{{S}} \right){a}=\left({m}_{{M}} −{m}_{{S}} \right){g} \\ $$$$\Rightarrow{a}=\frac{\left({m}_{{M}} −{m}_{{S}} \right){g}}{{m}_{{M}} +{m}_{{S}} } \\ $$
Commented by NECx last updated on 08/Dec/17
$${thanks}\:{boss}.\:{I}\:{now}\:{understand}\:{it}. \\ $$