A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. At the bottom, she is moving 20.3 m/s. How much work did the friction do on the woman?
+37084 Potential energy at the top of the slide (mgh) is converted to kinetic enegy at the bottom of the slide (1/2mv^2) ....the difference is the energy lost to friction
mgh = 77(9.81)(42.6 sin42.3) = 21656.71 j
1/2mv^2 = 1/2(77)(20.3^2) = 15865.47 j
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DIFFERENCE LOST TO FRICTION = 5791.25 j
+14985 A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. At the bottom, she is moving 20.3 m/s. How much work did the friction do on the woman?
\(h=42.6m\times sin42.3°\)
\(W_h = m \times g\times h =77kg\times\frac{9.81m}{s^2}\times42.6m\times sin42.3°\)
\(W_h= 21.656\ kNm\)
\(W_v= \frac{mv^2}{2}=\frac{77kg\times 20.3^2m^2}{2\times s^2}\)
\(W_v=15.865\ kNm\)
\(W_f=W_h-W_v=(21.656-15.865)kNm\)
\(W_f=5.791\ kNm\)
\(This \ is \ the \ friction \ work.\)
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