Menu Close

A-100kg-man-lowers-himself-to-the-ground-from-a-height-of-10m-by-means-of-a-rope-passed-over-a-frictionless-pulley-and-the-other-end-attached-to-a-70kg-sandbag-a-with-what-speed-does-the-man-hit-the




Question Number 25253 by NECx last updated on 07/Dec/17
A 100kg man lowers himself to  the ground from a height of 10m  by means of a rope passed over a  frictionless pulley and the other  end attached to a 70kg sandbag.  a)with what speed does the man   hit the ground?  b)could he have done anything to  reduce the speed?
$${A}\:\mathrm{100}{kg}\:{man}\:{lowers}\:{himself}\:{to} \\ $$$${the}\:{ground}\:{from}\:{a}\:{height}\:{of}\:\mathrm{10}{m} \\ $$$${by}\:{means}\:{of}\:{a}\:{rope}\:{passed}\:{over}\:{a} \\ $$$${frictionless}\:{pulley}\:{and}\:{the}\:{other} \\ $$$${end}\:{attached}\:{to}\:{a}\:\mathrm{70}{kg}\:{sandbag}. \\ $$$$\left.{a}\right){with}\:{what}\:{speed}\:{does}\:{the}\:{man}\: \\ $$$${hit}\:{the}\:{ground}? \\ $$$$\left.{b}\right){could}\:{he}\:{have}\:{done}\:{anything}\:{to} \\ $$$${reduce}\:{the}\:{speed}? \\ $$
Commented by NECx last updated on 07/Dec/17
please do well to interprete with a  diagram.That was 1 of my   challenge while I tried solving it.  Thanks
$${please}\:{do}\:{well}\:{to}\:{interprete}\:{with}\:{a} \\ $$$${diagram}.{That}\:{was}\:\mathrm{1}\:{of}\:{my}\: \\ $$$${challenge}\:{while}\:{I}\:{tried}\:{solving}\:{it}. \\ $$$${Thanks} \\ $$
Answered by mrW1 last updated on 07/Dec/17
a)  acceleration of man and sandbag is  a=(((m_M −m_S )g)/(m_M +m_S ))=((100−70)/(100+70))×10=((30)/(17)) m/s^2     speed of them is  v=(√(2ah))=(√(2×((30)/(17))×10))=5.96 m/s  b)  the man can reduce his speed if he also  holds the rope on the side of sandbay  and releases this rope gradually.
$$\left.{a}\right) \\ $$$${acceleration}\:{of}\:{man}\:{and}\:{sandbag}\:{is} \\ $$$${a}=\frac{\left({m}_{{M}} −{m}_{{S}} \right){g}}{{m}_{{M}} +{m}_{{S}} }=\frac{\mathrm{100}−\mathrm{70}}{\mathrm{100}+\mathrm{70}}×\mathrm{10}=\frac{\mathrm{30}}{\mathrm{17}}\:{m}/{s}^{\mathrm{2}} \\ $$$$ \\ $$$${speed}\:{of}\:{them}\:{is} \\ $$$${v}=\sqrt{\mathrm{2}{ah}}=\sqrt{\mathrm{2}×\frac{\mathrm{30}}{\mathrm{17}}×\mathrm{10}}=\mathrm{5}.\mathrm{96}\:{m}/{s} \\ $$$$\left.{b}\right) \\ $$$${the}\:{man}\:{can}\:{reduce}\:{his}\:{speed}\:{if}\:{he}\:{also} \\ $$$${holds}\:{the}\:{rope}\:{on}\:{the}\:{side}\:{of}\:{sandbay} \\ $$$${and}\:{releases}\:{this}\:{rope}\:{gradually}. \\ $$
Commented by NECx last updated on 07/Dec/17
how did you derive the formula  for the acceleration?
$${how}\:{did}\:{you}\:{derive}\:{the}\:{formula} \\ $$$${for}\:{the}\:{acceleration}? \\ $$
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=(m_(Man) −m_(Sand) )g,  therefore the acceleration of them  is a=(F/M)=(((m_M −m_S )g)/(m_M +m_S )).    An other way is:  let T=tension in rope  m_(Man) a=m_(Man) g−T    ...(i)  m_(Sand) a=T−m_(Sand) g   ...(ii)  (i)+(ii):  (m_M +m_S )a=(m_M −m_S )g  ⇒a=(((m_M −m_S )g)/(m_M +m_S ))
$${man}\:{and}\:{sandbag}\:{move}\:{together} \\ $$$${as}\:{a}\:{single}\:{object}.\:{total}\:{mass}\:{of}\:{them} \\ $$$${is}\:{M}={m}_{{Man}} +{m}_{{Sand}} .\:{The}\:{total}\:{force}\:{which} \\ $$$${acts}\:{on}\:{them}\:{is}\:{F}={m}_{{Man}} {g}−{m}_{{Sand}} {g}=\left({m}_{{Man}} −{m}_{{Sand}} \right){g}, \\ $$$${therefore}\:{the}\:{acceleration}\:{of}\:{them} \\ $$$${is}\:{a}=\frac{{F}}{{M}}=\frac{\left({m}_{{M}} −{m}_{{S}} \right){g}}{{m}_{{M}} +{m}_{{S}} }. \\ $$$$ \\ $$$${An}\:{other}\:{way}\:{is}: \\ $$$${let}\:{T}={tension}\:{in}\:{rope} \\ $$$${m}_{{Man}} {a}={m}_{{Man}} {g}−{T}\:\:\:\:…\left({i}\right) \\ $$$${m}_{{Sand}} {a}={T}−{m}_{{Sand}} {g}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\left({m}_{{M}} +{m}_{{S}} \right){a}=\left({m}_{{M}} −{m}_{{S}} \right){g} \\ $$$$\Rightarrow{a}=\frac{\left({m}_{{M}} −{m}_{{S}} \right){g}}{{m}_{{M}} +{m}_{{S}} } \\ $$
Commented by NECx last updated on 08/Dec/17
thanks boss. I now understand it.
$${thanks}\:{boss}.\:{I}\:{now}\:{understand}\:{it}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *