Question Number 47903 by Necxx last updated on 16/Nov/18
Commented by Necxx last updated on 16/Nov/18
$${mr}\:{Tanmay}\:{mrw}\:{and}\:{mr}\:{Ajfour} \\ $$$${please}\:{help} \\ $$
Commented by mr W last updated on 16/Nov/18
$${mass}\:{of}\:{the}\:{body}\:{must}\:{be}\:{given}\:{to} \\ $$$${get}\:{an}\:{answer}. \\ $$
Commented by Necxx last updated on 17/Nov/18
$${so}\:{if}\:{it}\:{were}\:{to}\:{be}\:{solved}\:{with}\:{the} \\ $$$${mass}\:{given}\:{as}\:{m}\:{what}\:{will}\:{be}\:{the} \\ $$$${procedure}? \\ $$
Commented by Necxx last updated on 17/Nov/18
$${ok}...{Thanks} \\ $$
Commented by mr W last updated on 17/Nov/18
$${assume}\:{the}\:{mass}\:{is}\:{m}=\mathrm{10}{kg}. \\ $$$${energy}\:{before}\:=\frac{{mv}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${energy}\:{lost}\:=\mathrm{105}\:{J} \\ $$$${energy}\:{after}={mgh} \\ $$$$\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{105}={mgh} \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{{g}}\left(\frac{{v}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{105}}{{m}}\right) \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{\mathrm{10}}\left(\frac{\mathrm{10}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{105}}{\mathrm{10}}\right)=\mathrm{3}.\mathrm{95}\:{m} \\ $$
Commented by Necxx last updated on 17/Nov/18
$${now}\:{I}\:{understand}.{Thank}\:{you}. \\ $$