A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of 1.10 m/s2? (c) Suppose you were per-forming the same experiment on the moon, where the acceleration due to gravity is 1.62 m/s2. (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

Solution 29E Mass of the crate m = 45.0 kg Weight of the crate, mg = 45.0 kg × 9.80 m/s 2 mg = 441 N Force to move the crate F = 313 N (a) Let, sbe the coefficient of static friction. Therefore, F = mgs 313 N = × s41 N = 313/441 s = 0.709 s When the crate is moving, the force on it is = 208 N Let ke the coefficient of kinetic friction. Therefore, 208 N = × k41 N k 208/441 = 0.471 k (b) When an acceleration is given to the crate, the force on the crate will be F = ma a 2 F a = 45.0 kg × 1.10 m/s F a = 49.5 N But the push that has to be given to the crate will also include the 208 N force to keep the crate moving at a constant speed. Therefore. total push F T = F a + 208 N F T = 49.5 N + 208 N F = 257.5 N T Therefore, the total push needed to accelerate the crate is 257.5 N. 2 (c) i) Given that, moon = 1.62 m/s The force of static friction is smoon= 0.709 × 45 × 1.62 N F smoon= 51.6 N Therefore, the force required to move the crate on the moon is 51.6 N. (ii) The push force from part b is FT = 257.5 N Therefore, F = 45 × a moon+ mg moon T k 257.5 N = 45a + 0.471 × 45 × 1.62 N moon 257.5 N = 45a moon+ 34.3 N 45a moon= 223.2 N 2 a moon = 223.2/45 m/s a moon = 4.96 m/s 2 Therefore, the acceleration of the crate on the moon will be 4.96 m/s .