A particle $$\(P\)$$ of mass $$\(0.3 \mathrm{~kg}\)$$ rests on a rough plane inclined at an angle $$\(\theta\)$$ to the horizontal, where $$\(\sin \theta=\frac{7}{25}\)$$. A horizontal force of magnitude $$\(4 \mathrm{~N}\)$$, acting in the vertical plane containing a line of greatest slope of the plane, is applied to $$\(P\)$$ (see diagram). The particle is on the point of sliding up the plane. The force acting horizontally is replaced by a force of magnitude 4 N acting up the plane parallel to a line of greatest slope. Starting with $$\(P\)$$ at rest, the force of $$\(4 \mathrm{~N}\)$$ parallel to the plane acts for 3 seconds and is then removed. Find the total distance travelled until $$\(P\)$$ comes to instantaneous rest. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_s21_qp_43 Year：2021 Question No：7(c)
Answer:
\( s_{1}=\frac{1}{2} \times \frac{10}{3} \times 3^{2}=15 \text { and } v=\frac{10}{3} \times 3=10
\)
\(-0.3 g \times \sin \theta-\mu \times 0.3 g \cos \theta=0.3 a\) leading to \(a=-10\)
\( 0=10^{2}+2 \times(-10) \times s_{2}
\)
\(\left[s_{2}=5\right.\) leading to total distance \(\left.=s_{1}+s_{2}=15+5=\right] 20 \mathrm{~m}\)
Alternative method for Question 7(c)
Work done \(=4 \times 0.5 \times \frac{10}{3} \times 3^{2}[=60 \mathrm{~J}]\)
\( 60=\mu \times 0.3 g \cos \theta \times d+0.3 g \times d \sin \theta
\)
\(d=20 \mathrm{~m}\)
\)
\(-0.3 g \times \sin \theta-\mu \times 0.3 g \cos \theta=0.3 a\) leading to \(a=-10\)
\( 0=10^{2}+2 \times(-10) \times s_{2}
\)
\(\left[s_{2}=5\right.\) leading to total distance \(\left.=s_{1}+s_{2}=15+5=\right] 20 \mathrm{~m}\)
Alternative method for Question 7(c)
Work done \(=4 \times 0.5 \times \frac{10}{3} \times 3^{2}[=60 \mathrm{~J}]\)
\( 60=\mu \times 0.3 g \cos \theta \times d+0.3 g \times d \sin \theta
\)
\(d=20 \mathrm{~m}\)
Knowledge points:
4.1.6 understand the concepts of limiting friction and limiting equilibrium, recall the definition of coefficient of friction, and use the relationship F = nR or F G nR, as appropriate
4.2.4 use appropriate formulae for motion with constant acceleration in a straight line. （Questions may involve setting up more than one equation, using information about the motion of different particles.）
4.4.1 apply Newton’s laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces, which may include friction, tension in an inextensible string and thrust in a connecting rod If any other forces resisting motion are to be considered (e.g. air resistance) this will be indicated in the question.
4.4.3 solve simple problems which may be modelled as the motion of a particle moving vertically or on an inclined plane with constant acceleration Including, for example, motion of a particle on a rough plane where the acceleration while moving up the plane is different from the acceleration while moving down the plane.
Solution:
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