Hi I need help with the following question:

First attempt:

Suppse the box coverts its kinetic energy \(\displaystyle W_{k} \) into work of friction \(\displaystyle W_{f} \) where \(\displaystyle W_{f} = F_{k}S \) and \(\displaystyle F_{k} \) denotes the force of kinetic friction and \(\displaystyle S \) the displacement. Then it follows that

\(\displaystyle W_{k} = W_{f} \iff (1/2)(mv^2)=F_{k}S \iff F_{k} = \frac{mv^2}{2S}\approx 39,2 N\) . We know that \(\displaystyle \mu_{k}= \frac {F_{k}}{F_{N}} \) and by trigonomety \(\displaystyle F_{N}=-mgcos(20^{\circ}) \) therefore

\(\displaystyle \mu_{k} = \frac{\frac{mv^2}{2S}}{-mgcos(20^{\circ})}=\frac{mv^2}{-2Smgcos(20^{\circ})}=\frac{v^2}{-2Sgcos(20^{\circ})}=\frac{(5.6)^{2}}{4(9.82)cos(20^{\circ})}\approx 0.8496\)

Second attempt:

The resultant force \(\displaystyle F_{∥ }\) is parallel to the contact surface. We know that the sum of forces acting on the box parallel to the contact surface is equal to \(\displaystyle F_{∥ } =Fg_{∥} + F_{k} \) where \(\displaystyle Fg_{∥} \) denotes the component force of gravity with is parellel to the contact surface and\(\displaystyle F_{k} \) the force of kinetic friction. From trigonomety \(\displaystyle Fg_{∥}=Sin(20^{\circ}) Fg=Sin(20^{\circ}) mg\) and the normal force \(\displaystyle F_{N}=cos (20^{\circ})mg\). By definition \(\displaystyle F_{k} =\mu_{k}(F_{N}) =\mu_{k}cos (20^{\circ})mg\)

Since \(\displaystyle F_{∥ }=Fg_{∥} + F_{k} =Sin(20^{\circ})mg +\mu_{k}Cos(20^{\circ})mg = mg(Sin(20^{\circ})+\mu_{k}Cos(20^{\circ}))=a(m)\). Then the acceleration of the box is equal to

\(\displaystyle a = g(Sin(20^{\circ})+\mu_{k}Cos(20^{\circ}))\). From kinematics we can calculate \(\displaystyle a \) by using the formula \(\displaystyle v_{f}^{2}=v_{i}^{2}+2as\) given that \(\displaystyle v_{i}=5.6 \space m/s \), \(\displaystyle v_{f}=0 \space m/s \) and \(\displaystyle s =2 \space m\). Then \(\displaystyle a = -\frac{v_{i}^{2}}{2s}= -\frac{5.6^{2}}{4}=-7.84\). Now we can find \(\displaystyle \mu_{k} \) given that \(\displaystyle g=-9.82 \space m/s^{2} \) thus \(\displaystyle \mu_{k} =\frac{a-gSin(20^{\circ})}{gCos(20^{\circ})} \approx 0.4856 \)

**A small box with the mass of 5 kg is sent with an inital speed of 5.6 m/s up an sloping plane with an incline angle of 20 °. After 2.0 m the box stops. Determine the coefficient of friction between the box and the plane.**First attempt:

Suppse the box coverts its kinetic energy \(\displaystyle W_{k} \) into work of friction \(\displaystyle W_{f} \) where \(\displaystyle W_{f} = F_{k}S \) and \(\displaystyle F_{k} \) denotes the force of kinetic friction and \(\displaystyle S \) the displacement. Then it follows that

\(\displaystyle W_{k} = W_{f} \iff (1/2)(mv^2)=F_{k}S \iff F_{k} = \frac{mv^2}{2S}\approx 39,2 N\) . We know that \(\displaystyle \mu_{k}= \frac {F_{k}}{F_{N}} \) and by trigonomety \(\displaystyle F_{N}=-mgcos(20^{\circ}) \) therefore

\(\displaystyle \mu_{k} = \frac{\frac{mv^2}{2S}}{-mgcos(20^{\circ})}=\frac{mv^2}{-2Smgcos(20^{\circ})}=\frac{v^2}{-2Sgcos(20^{\circ})}=\frac{(5.6)^{2}}{4(9.82)cos(20^{\circ})}\approx 0.8496\)

Second attempt:

The resultant force \(\displaystyle F_{∥ }\) is parallel to the contact surface. We know that the sum of forces acting on the box parallel to the contact surface is equal to \(\displaystyle F_{∥ } =Fg_{∥} + F_{k} \) where \(\displaystyle Fg_{∥} \) denotes the component force of gravity with is parellel to the contact surface and\(\displaystyle F_{k} \) the force of kinetic friction. From trigonomety \(\displaystyle Fg_{∥}=Sin(20^{\circ}) Fg=Sin(20^{\circ}) mg\) and the normal force \(\displaystyle F_{N}=cos (20^{\circ})mg\). By definition \(\displaystyle F_{k} =\mu_{k}(F_{N}) =\mu_{k}cos (20^{\circ})mg\)

Since \(\displaystyle F_{∥ }=Fg_{∥} + F_{k} =Sin(20^{\circ})mg +\mu_{k}Cos(20^{\circ})mg = mg(Sin(20^{\circ})+\mu_{k}Cos(20^{\circ}))=a(m)\). Then the acceleration of the box is equal to

\(\displaystyle a = g(Sin(20^{\circ})+\mu_{k}Cos(20^{\circ}))\). From kinematics we can calculate \(\displaystyle a \) by using the formula \(\displaystyle v_{f}^{2}=v_{i}^{2}+2as\) given that \(\displaystyle v_{i}=5.6 \space m/s \), \(\displaystyle v_{f}=0 \space m/s \) and \(\displaystyle s =2 \space m\). Then \(\displaystyle a = -\frac{v_{i}^{2}}{2s}= -\frac{5.6^{2}}{4}=-7.84\). Now we can find \(\displaystyle \mu_{k} \) given that \(\displaystyle g=-9.82 \space m/s^{2} \) thus \(\displaystyle \mu_{k} =\frac{a-gSin(20^{\circ})}{gCos(20^{\circ})} \approx 0.4856 \)

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