A car of mass $$\(1500 \mathrm{~kg}\)$$ is pulling a trailer of mass $$\(750 \mathrm{~kg}\)$$ up a straight hill of length $$\(800 \mathrm{~m}\)$$ inclined at an angle of $$\(\sin ^{-1} 0.08\)$$ to the horizontal. The resistances to the motion of the car and trailer are $$\(400 \mathrm{~N}\)$$ and $$\(200 \mathrm{~N}\)$$ respectively. The car and trailer are connected by a light rigid tow-bar. The car and trailer have speed $$\(30 \mathrm{~m} \mathrm{~s}^{-1}\)$$ at the bottom of the hill and $$\(20 \mathrm{~m} \mathrm{~s}^{-1}\)$$ at the top of the hill. Use an energy method to find the constant driving force as the car and trailer travel up the hill. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_w20_qp_41 Year：2020 Question No：6(a)
Answer:
\(\mathrm{KE}(\) final \()=1 / 2 \times 1500 \times 20^{2}+1 / 2 \times 750 \times 20^{2}\)
\(\mathrm{KE}(\) initial \()=1 / 2 \times 1500 \times 30^{2}+1 / 2 \times 750 \times 30^{2}\)
PE gain \(=2250 \times 10 \times 800 \times 0.08\)
WD against friction \(=600 \times 800\)
\(1 / 2 \times 2250 \times 30^{2}+\mathrm{DF} \times 800=600 \times 800\)
\(+1 / 2 \times 2250 \times 20^{2}+2250 \times 10 \times 800 \times 0.08\)
\(\mathrm{DF}=1700 \mathrm{~N}\)
\(\mathrm{KE}(\) initial \()=1 / 2 \times 1500 \times 30^{2}+1 / 2 \times 750 \times 30^{2}\)
PE gain \(=2250 \times 10 \times 800 \times 0.08\)
WD against friction \(=600 \times 800\)
\(1 / 2 \times 2250 \times 30^{2}+\mathrm{DF} \times 800=600 \times 800\)
\(+1 / 2 \times 2250 \times 20^{2}+2250 \times 10 \times 800 \times 0.08\)
\(\mathrm{DF}=1700 \mathrm{~N}\)
Knowledge points:
4.5.1 understand the concept of the work done by a force, and calculate the work done by a constant force when its point of application undergoes a displacement not necessarily parallel to the force W = Fd cos; Use of the scalar product is not required.
4.5.2 understand the concepts of gravitational potential energy and kinetic energy, and use appropriate formulae
4.5.3 understand and use the relationship between the change in energy of a system and the work done by the external forces, and use in appropriate cases the principle of conservation of energy Including cases where the motion may not be linear (e.g. a child on a smooth curved ‘slide’), where only overall energy changes need to be considered.
Solution:
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