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Question Number 69829 by TawaTawa last updated on 28/Sep/19

Commented by TawaTawa last updated on 28/Sep/19

$$\mathrm{The}\mathrm{question}\mathrm{says}\mathrm{do}\mathrm{not}\mathrm{use}{\mathrm{Newton}}^{\prime }\mathrm{s}\mathrm{law}\mathrm{and}\mathrm{kinematic}$$

Commented by TawaTawa last updated on 28/Sep/19

$$\mathrm{Help}\mathrm{with}\mathrm{the}\left(\mathrm{b}\right)\mathrm{sir}$$

Answered by mr W last updated on 28/Sep/19

$$\left(b\right)$$$$energyatthebeginning:$$$$E_1=\frac{1}{2}{mv}^2$$$$energyattheend:$$$$E_2=mgh$$$$workdonebythegravityforce:$$$$-mgh$$$$frictionforce:$$$$f=\mu N=\mu mg\cos \theta$$$$workdonebythefrictionforce:$$$$-fs=-f\frac{h}{\sin \theta }=-\frac{\mu mgh}{\tan \theta }$$$$work=energychange$$$$-mgh-\frac{\mu mgh}{\tan \theta }=mgh-\frac{1}{2}{mv}^2$$$$\Rightarrow \mu =(2-\frac{v^2}{2gh})\tan \theta =(2-\frac{{12}^2}{2\times 10\times 5})\times \tan 45°=0.56$$

Commented by TawaTawa last updated on 28/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 28/Sep/19

$$\left(a\right)$$$$workdonebythepullingforce:$$$$F\cos \theta s=20\times \frac{4}{5}\times 5=80J$$$$frictionforce:$$$$f=\mu N=\mu (mg-F\sin \theta )=0.1(10\times 10-20\times \frac{3}{5})=8.8N$$$$workdonebythefrictionforce:$$$$fs=-8.8\times 5=-44J$$$$workdonebyallforces:$$$$W=80-44=36J$$$$kineticenergyofmass:$$$$E=\frac{1}{2}{mv}^2$$$$work=energychange:$$$$W=E$$$$36=\frac{1}{2}\times 10\times v^2$$$$\Rightarrow v=\sqrt{\frac{36}{5}}=\sqrt{7.2}=2.68m/s$$

Commented by TawaTawa last updated on 28/Sep/19

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}$$

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