physics
posted by Keelia on .
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 blockfloor system, and (c) the coefficient of kinetic friction between the block and floor?

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. 
Correction:
u = Ff / (FvFap*sin29.8),
u = 5.09 / (24.895.87sin29.8) = 0.23.