BASIC PHYSICS FOR MANAJJEMENT INFORMATICS AND COMPUTER ENGINEERING

BASIC PHYSICS FOR MANAJJEMENT INFORMATICS AND COMPUTER ENGINEERING
UNIVERSITY GUNADARMA
DEPOK WEST JAVA
BY RADEN BAYU Herlambang

Applied Natural Science Quiz 1:

Do the following questions!

1. A stone fell from the building that y-meter height above the ground. Speed rock as it reached the ground is 20 m / s. If g = 10 m/s2, determine the value of y!

2. A bucket of sand hanging by a rope has mass of 5 kg. If the acceleration of gravity of the earth in that place (g = 10 m/s2), then calculate the magnitude of the force strap tension,

If:

a. system is at rest

b. system is moving upward with acceleration of 2 m/s2.

3. At the point O (x-axis intersection point and the y-axis) worked three styles in a single plane that is: F1 = 4 N, F2 = 3 N and F3 = 8 N. Each of these styles in a row forming an angle of 0o, 90o and 217o to the positive x-axis. Determine the resultant force of the three great style.

4. A ball kicked by Tambunan with elevation angle 60 ° (sin 60 ° = 0.87; cos 60 ° = 0.5) with initial velocity of 30 m / s. If the acceleration of gravity (g = 10m/s2), then specify:

a. position at t = 1 s,

b. velocity at t = 1 s,

c. The highest point coordinates,

5. Determine the energy required to lift beam 10 kg from ground level to the table at 1.5 m. The acceleration of gravity g = 10 m/s2.

1. A stone fell from the top of a tall building h feet above

soil. Speed rock as it reached the ground = 20 m / s. If g = 10 m/s2,

specify the value H!

Completion:

Given: vt = 20 m / s

g = 10 m/s2

Asked: h

Answer:

Vt = vo + g. t

20 = 0 + 10. t

t = 2 seconds

2h = 4 .10

2h = 40

h = 20 m

2. A bucket of sand hanging by a rope has mass 5 Kg. If the acceleration of gravity of the earth in a place that g = 10 m/s2, then calculate

large rope tension force if:

a. system is at rest

b. system is moving upward with acceleration of 2 m/s2.

a. Because the system is silent, then apply HK. I Newton

ΣF = 0

T - W = 0

T = W = m. g

T = 5. 10 = 50 N

b. Because the system is moving upward, caused by

constant force, then apply Newton HK.II.

ΣF = m. a

T - W = m. a

T = m. a + W

T = 5. 2 + 50 = 60 N

3. At the point O (x-axis intersection point and the y-axis) worked on three styles

one plane are: F1 = 4 N, F2 = 3 N and F3 = 8 N. Each style

those on the positive x-axis respectively form an angle 0o, 90o

and 217o. Determine the resultant force from the three major styles

Completion:

Given: F1 = 4 N, F2 = 3 N; F3 = 8 N

α1 = 0o; α2 = 90o; α3 = 217o

Asked: R = ...?

Answer:

a.

= 3 N

4. A ball kicked by Tambunan with elevation angle 60 ° (sin 60 ° = 0.87 and cos 60 ° = 0.5) with initial velocity of 30 m / s. If the acceleration of gravity g = 10 m/s2, determine:

a. position at t = 1 s,

b. velocity at t = 1 s,

c. The highest point coordinates,

Completion:

Given: θ = 60 °

Vo = 30 m / s

Asked: a. (X, y) t = 1 =. . . ?

d. vt = 1 =. . . ?

c. (X, y) = highest. . . ?

Answer:

a. x = vo. cos θ. t = 30. cos 60 °. 2 = 30. 0.5. 1 = 15 m

y = vo. sin θ. t - ½. g. t 2

= 30. sin 60 °. 2 - ½. 10. (1) 2

= 30. 0.87 .1 - 5. 1

= 26.1-5

= 21.1 m

So, the position of the ball after 1 s is (20, 21.1).

b. vx = vo. cos θ = 30. cos 60 ° = 30. 0.5 = 15 m / s

vy = vo. sin θ - g. t = 30. sin 60 ° - 10. 1 = 30. 0.87-10 = 16.1 m / s

c. At the highest point vy = 0, so that:

vy = vo. sin θ - g. t

0 = 30. sin 60 ° - 10. t

0 = 26.1 to 10. t

10. t = 26.1

t = 2.61 s

T value is then entered on

x = vo. cos θ. t = 30. cos 60 °. = 10 m

y = vo. sin θ. t - ½. g. t2

= 30. sin 60 °. 2.61 - ½. 10. (2,61) 2

= 68.1 to 34

= 34.1 m / s

So, position the ball at the highest point is (10,34,1) m / s.

5. Determine the energy required to lift beam 10 kg from ground level to the table at 1.5 m. The acceleration of gravity g = 10 m/s2.

Completion:

Given:

m = 10 kg

h1 = 0 m

h2 = 1.5 m

Asked: W =. . .?

Answer:

W = m. g (h1 - h2)

W = 10. 10. (0 - 1.5)

W = - 150 joules

Sign (-) means an amount of energy required to lift the beam.

Sitompul riding a motorcycle racing with an acceleration of 4 m/s2. Find the speed Sitompul after bergerakselama 10 seconds, if the initial speed of zero!

vt = vo + a. t

= 0 + 4. 10

= 40 m / s