A slide in a playground descends at a constant angle of $$$30^{\circ}$$$ for $$$2.5 \mathrm{~m}$$$. It then has a horizontal section in the same vertical plane as the sloping section. A child of mass $$$35 \mathrm{~kg}$$$, modelled as a particle $$$P$$$, starts from rest at the top of the slide and slides straight down the sloping section. She then continues along the horizontal section until she comes to rest (see diagram). There is no instantaneous change in speed when the child goes from the sloping section to the horizontal section. The child experiences a resistance force on the horizontal section of the slide, and the work done against the resistance force on the horizontal section of the slide is $$$250 \mathrm{~J}$$$ per metre. It is given that the sloping section of the slide is smooth. Find the distance that the child travels along the horizontal section of the slide before she comes to rest. ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................

Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_s21_qp_41 Year：2021 Question No：7(a)(ii)

Answer:

$ \frac{1}{2} \times 35 \times 5^{2}=250 d
d=1.75 \mathrm{~m} $
Alternative method for Question 7(a)(ii)
$ 35 g \times 2.5 \sin 30=250 d $
$ d=1.75 \mathrm{~m} $
Alternative method for Question 7(a)(ii)
$-250=35 a$ leading to $a=-\frac{50}{7}=-7.14$
$0=5^{2}+2(a) d$
$d=1.75 \mathrm{~m}$

Knowledge points:

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.

Answer:

Knowledge points:

Solution:

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