Page 26 - Physics Form 5 KSSM_Neat

Solution

(a) Step 1: Step 2: Step 3: Step 4:
Identify the Identify the Identify the formula Solve the problem
problem information given that can be used numerically

KEMENTERIAN PENDIDIKAN MALAYSIA
 Magnitude of the horizontal component, T  T = 36 cos 30 o
x
x
and vertical component, T of the pull, T = 31.18 N (to the right) T y T
y
T = 36 sin 30 o
 Angle above the horizontal surface = 30° = 18.00 N (upwards) 30°
y
Magnitude of the pulling force, T = 36 N T x
 T = T cos 30 o
x
T = T sin 30 o
y
(b) Horizontal component to the right, (c) Resultant force, F = 11.18 N
T = 31.18 N Mass of block, m = 2.4 kg
x
Frictional force, F = 20 N F = ma
R
Resultant of horizontal components F
= T + F y Acceleration of the block, a = m
x
= 31.18 + (–20) = 11.18
= 11.18 N 2.4
= 4.66 m s –2
Vertical component upwards, T = 18.00 N
y
Normal reaction, R = 6 N
Weight, W = 24 N
Resultant of vertical components
= T + R + W
y
= 18 + 6 + (–24)
= 0 N
Resultant force on the block, F is 11.18 N
to the right.

Example 2

Figure 1.20 shows the free body diagram of a block sliding down a
smooth inclined plane. Normal reaction = 12 N
(a) Sketch the component of the weight of the block parallel to the
inclined plane and the component of the weight of the block Block
perpendicular to the inclined plane.
(b) Determine the resultant force acting on the block.
(c) Calculate the acceleration of the block if its mass is 2.4 kg.
Weight,
W = 24 N 60°
Figure 1.20

16 LS 1.2.2