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A 26-kg sled is on a snow-covered slope. The coefficients of friction between the sled’s runners and the snow are µs = 0.096 and µk = 0.072. What pulling force must be exerted on the sled as pictured in order to (a) start the sled sliding and (b) continue to slide the sled at constant velocity?

User WebBrother
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1 Answer

5 votes

Answer:

Step-by-step explanation:

Given

mass of sled =26 kg

coefficient of static friction
\mu _s=0.096

coefficient of kinetic friction
\mu _k=0.072

In order to move sled from rest we need to provide a force greater than static friction which is given by


f_s=\mu mg=0.096* 26* 9.8=24.46 N

After Moving Sled kinetic friction comes in to play which is less than static friction


f_k=\mu _kmg=0.072* 26* 9.8=18.34 N

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

User Benny Hallett
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