Asked by Keelia
A rope is used to pull a 2.54 kg block at constant speed 4.48 m along a horizontal floor. The force on the block from the rope is 5.87 N and directed 29.8° above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?
Answers
Answered by
Henry
Wb = mg = 2.54kg * 9.8N/kg = 24.89N. =
Weight of block.
Fb = 4.89N @ 0deg.
Fp = 24.89sin(0) = 0 = Force parallel to floor.
Fv = 24.89cos(0) = 24.89N. = Force perpendicular to floor.
a. W = Fap*cos29.8 * d,
W = 5.87cos29.8 * 4.48 = 5.1J.
c. Fn = Fap*cos29.8 - Fp - Ff = 0,
5.87cos29.6 - 0 - Ff = 0,
Ff = 5.87cos29.8 = 5.09N. = Force of friction.
u = Ff / Fv = 5.09 / 24.89 = 0.205.
Weight of block.
Fb = 4.89N @ 0deg.
Fp = 24.89sin(0) = 0 = Force parallel to floor.
Fv = 24.89cos(0) = 24.89N. = Force perpendicular to floor.
a. W = Fap*cos29.8 * d,
W = 5.87cos29.8 * 4.48 = 5.1J.
c. Fn = Fap*cos29.8 - Fp - Ff = 0,
5.87cos29.6 - 0 - Ff = 0,
Ff = 5.87cos29.8 = 5.09N. = Force of friction.
u = Ff / Fv = 5.09 / 24.89 = 0.205.
Answered by
Henry
Correction:
u = Ff / (Fv-Fap*sin29.8),
u = 5.09 / (24.89-5.87sin29.8) = 0.23.
u = Ff / (Fv-Fap*sin29.8),
u = 5.09 / (24.89-5.87sin29.8) = 0.23.
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