Answer:
the work done by the 30N force is 4156.92 J.
For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:
W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J
Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822
Answer:CCCCCCCCCCCCCCCCC
Explanation:CCCCCCCCCCCCC
Answer:
ill help if u help me?its 4am
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