1. This document discusses concepts of physics related to acceleration, including calculations of net force, acceleration, displacement, time, and tension. It provides 10 examples of word problems and their step-by-step solutions involving these concepts. 2. The examples calculate values such as net force, acceleration, displacement, time, tension and maximum static friction based on given values of mass, force, weight, velocity, displacement and time. 3. Formulas used include calculations for net force, acceleration, velocity, displacement, time, tension, and static friction.

1. FISIKA
Kartika Michele Nalle
X AKSELERASI

2. 1. Dik: W = 100 N
θ = 30o
F = 40 N
Dit : N?
Jwb :
N = W - F sinθ
= 100N - 40 N. 1/2
= 100 N – 20 N
= 80 N

3. 2. Dik: m =1kg
s = 2m
v = 4m/s
Dit : F?
Jwb : v2 = 2.a.s
(4m/s)2 = 2. a. 2m
16m2/s2 = 4m.a
a = 4 m/s2
F = m.a
= 1kg.4m/s2
= 4N

4. 3. Dik: m = 5 kg
F = 25 N
Dit: a = ..?
perpindahan balok setelah2 sekon = ..?
Jwb : F = m.a
25 N = 5 kg . a
a = 25 kg m s-2 : 5 kg
= 5 m s-2
Perpindahansetelah2 sekon:
2 s . 5 m s-2 = 10 m

5. 4. Dik: m = 4kg
t = 3s
s = 2,25 m
Dit : F?
Jwb : s = ½ a.t2
2 ¼ m = ½ a.9s2
2 ¼m . 2/9s2 = a
½ m/s2 = a
F = m.a
= 4kg . ½ ms-2
= 2N

6. 5. Dik : m = m
F = F
a = a
Dit : Jikam = 2m, dan F = 4F, maka a = ..?
Jwb : F = m . a
4F = 2m.a
a = 4F/2m
= 2a

7. 6. Dik: m1 = 5.000 kg
t = 50 s
v = 72 km/jam = 20 m/s
Dit : F ?
Jwb : a = v/t
= 20ms-1 /50 s
= 2/5 ms-2
F = m.a
= 5.000 kg . 2/5 ms-2
= 2000 N

8. 7. Dik : m = 10kg
a = 5m/s2
g = 10 m/s2
Dit : T?
Jwb : T = m.g – m.a
= 10kg. 10m/s2 – 10kg. 5m/s2
= 100 N – 50 N
= 50 N

9. 8. Dik : m = 40kg
a = 5m/s2
g = 10 m/s2
Dit : T?
Jwb : T = m.g+ m.a
= 40kg. 10m/s2 – 40kg. 5m/s2
= 400 N + 200 N
= 600 N

10. 9. Dik: m1=1 kg
m2 =2 kg
g = 10m/s2
Dit:a ?
T ?
Jwb: a=(m2 – m1 /m2 +m1) g
=( 2kg-1kg/2kg+ 1kg)10m/s2
= (1kg/3kg)10m/s2
=10/3 m/s2 atau3,3m/s2
T= m2.g– m2.a
= 2kg. 10 m/s2 -2kg.10/3 m/s2
= 20 N– 20/3N
=40/3 N =13,33333N

11. 10..Dik:m1 =1 kg
m2 = 4kg
fs1max = 4N
Dit: fs max?
a?
Jwb : a. m1 : m2 = 1 : 4
fs1max : fs2max = 4N:x
x = 16N
fmax = fs1 max +f s2 max
fmax= 4n +16 N
= 20N
b.F =m.a
• 20N = (1kg +4kg) .a
a=20N / 5kg
a =4m/s2