A 310 kg crate hangs from the end of a rope of length L = 11.6 m. You push horiz

Question

A 310 kg crate hangs from the end of a rope of length L = 11.6 m. You push horizontally on the crate with a varying force to move it distance d = 3.35 m to the side. (a) What is the magnitude of when the crate is in this final position? During the crate's displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (d) the work done by the pull on the crate from the rope? (e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force does on the crate.

Explanation / Answer

A 300 kg crate hangs from the end of a rope of length L=11.6 m. You push horizontally on the crate with a varying force F to move it distance d=3.35 m to the side the product of the horizontal displacement and the answer to (a)? Solution: (a) To hold the crate at equilibrium in the final situation, r F must have the same magnitude as the horizontal component of the rope’s tension T sin q , where q is the angle between the rope (in the final position) and vertical: = = 0.333? = 19.47° 12 4 sinq q But the vertical component of the tension supports against the weight: T cos q = mg . Thus, the tension is T = (300 kg)(9.80 m/s2)/cos 19.5° = 3118.9 N and F = (3118.9 N) sin 19.5° = 1041 N. 7 An alternative approach based on drawing a vector triangle (of forces) in the final situation provides a quick solution. (b) Since there is no change in kinetic energy, the net work on it is zero. (c) The work done by gravity is W F d mgh g g = × = - r r , where h = L(1 – cos q ) is the vertical component of the displacement. With L = 11.06m, we obtain Wg = –164 J which should be rounded to three figures: –1.64 kJ. (d) The tension vector is everywhere perpendicular to the direction of motion, so its work is zero (since cos 90° = 0). (e) The implication of the previous three parts is that the work due to r F is –Wg (so the net work turns out to be zero). Thus, WF = –Wg = 1.64 kJ.

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