Page 27 - Physics Form 5 KSSM_Neat

Solution CHAPTER 1
Normal reaction = 12 N
(a) Figure 1.21 shows a sketch of the component of the weight
of the block parallel to the inclined plane, W and the W y Force and Motion II
x
component of the weight of the block perpendicular to the 60°
inclined plane, W . W x
y
KEMENTERIAN PENDIDIKAN MALAYSIA
(b) W = 24 sin 60°
x
= 20.78 N
W = 24 cos 60° 60°
y
= 12 N Figure 1.21
Resultant of the forces perpendicular to the Info
inclined plane = 12 + (–12)
= 0 N Resolution of forces for object on
Resultant force on the block = 20.78 N inclined plane
Given ˙ABC = θ
(c) Resultant force, F = 20.78 N Thus, ˙BDE = 90° – θ
Mass of block, m = 2.4 kg and ˙EDF = 90° – (90° – θ)
= θ
F = ma C
F
Acceleration of block, a =
m
20.78
=
2.4 D
= 8.66 m s –2
90° – θ
θ
F
θ
Formative Practice 1.2 A E B

1. Resolve the following forces into horizontal component and vertical component.
(a) (b)
64°
70 N
90 N
42°
2. Figure 1.22 shows a man pushing a lawn mower with a force of 90 N.

90 N
60°

Figure 1.22
(a) Resolve the pushing force into its horizontal component and vertical component.
(b) State the function of the horizontal component and vertical component of the pushing
force when the lawn mower is being pushed.

LS 1.2.2 17