Answer:
(A). The work done by friction in crossing the patch is -637.98 J.
(B). The speed of skier is 10.57 m/s.
Explanation:
Given that,
Mass of skier = 62 kg
Speed = 6.5 m/s
Length = 3.50 m
Coefficient kinetic friction = 0.30
Height = 2.5 m
(A) we need to calculate the work done by friction in crossing the patch
Using formula of work done
![W=-\mu mg\times l](https://tex.z-dn.net/?f=W%3D-%5Cmu%20mg%5Ctimes%20l)
Put the value into the formula
![W=-0.30\times62\times9.8\times3.50](https://tex.z-dn.net/?f=W%3D-0.30%5Ctimes62%5Ctimes9.8%5Ctimes3.50)
![W=-637.98\ J](https://tex.z-dn.net/?f=W%3D-637.98%5C%20J)
The work done by friction in crossing the patch is -637.98 J.
(B) we need to calculate the speed of skier
Using conservation of energy
![K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}](https://tex.z-dn.net/?f=K.E_%7Bi%7D%2BU_%7Bi%7D-W_%7Bfriction%7D%3DK.E_%7Bf%7D%2BU_%7Bf%7D)
![\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv_%7B1%7D%5E2%2Bmgh-%5Cmu%20mgl%3D%5Cdfrac%7B1%7D%7B2%7Dmv_%7B2%7D%5E2%2BU_%7Bf%7D)
Final potential energy is zero
So, ![\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv_%7B1%7D%5E2%2Bmgh-%5Cmu%20mgl%3D%5Cdfrac%7B1%7D%7B2%7Dmv_%7B2%7D%5E2)
![\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dv_%7B2%7D%5E2%3D%5Cdfrac%7B1%7D%7B2%7Dv_%7B1%7D%5E2%2Bgh-%5Cmu%20gl)
Put the value into the formula
![\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dv_%7B2%7D%5E2%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes6.5%5E2%2B9.8%5Ctimes2.5%2B0.30%5Ctimes9.8%5Ctimes3.50)
![v_{2}=\sqrt{2\times55.915}](https://tex.z-dn.net/?f=v_%7B2%7D%3D%5Csqrt%7B2%5Ctimes55.915%7D)
![v_{2}=10.57\ m/s](https://tex.z-dn.net/?f=v_%7B2%7D%3D10.57%5C%20m%2Fs)
The speed of skier is 10.57 m/s.
Hence, (A).The work done by friction in crossing the patch is -637.98 J.
(B).The speed of skier is 10.57 m/s.