A block of mass 500 g is pulled from rest on a horizontal frictionless bench by a steady force F and travels 8 m in 2 s. Find a the acceleration, b the value of F.
1. A block of mass 500 g is pulled from rest on a horizontal frictionless bench by a steady force F and travels 8 m in 2 s. Find a the acceleration, b the value of F.
Jawaban:
F = m × a
d = vi × t + ½ × a × t²
Variabel yang Diberikan
m = 5 kg
vi = 0 m/s
d = 8 m
t = 2 s
8 = 0 m/s × 2s + ½ × a × 2s²
a = 4 m/s²
F = 5 kg × 4 m/s
F = 2 N
2. A bullet of mass 0.20 kg travels horizontally at a speed of 400 m/s. it strikes a stationary wooden block of mass 1.80 kg resting on a frictionless, horizontal surface. the bullet stays in the block. what is the speed of the bullet and the block immediately after the impact? 200 m/s 80 m/s 44 m/s 40 m/s (in ulanganku hari ini)
Jawaban:
40 m/s
Penjelasan:
Known :
Mb = 0.20 kg
Vb = 400 m/s
Mw = 1.8 kg
Vw = 0 m/s (Resting)
Solution :
Because bullet stays in the block, then the speed of the bullet and the block will be same :
Mb . Vb + Mw . Vw = (Mb + Mw) . V
0.2 . 400 + 0 = (0.2 + 1.8) . V
V = 40 m/s
3. A block of mass 5 kg initially at rest at the origin is acted on by a force along the positive X - direction represented by F= (20 +5x) N. Calculate the work done by the force during the displacement of the block from x=0 at x = 4mMakasih ^^"
Question:
A block of mass 5 kg initially at rest at the origin is acted on by a force along the positive X - direction represented by F= (20 +5x) N. Calculate the work done by the force during the displacement of the block from x=0 at x = 4mAnswer:
The work done by the body is 10 JoulesExplanation :
Given that :
A block of mass 5 kg initially is at rest Aforce of F = ( 20 + 5x ) N is acted on it along the positive x direction along the displacement of x = 0 to x = 4mToFind:
The work done on the bodyFormula Used :
[tex]\bigstar \; {\underline{\boxed{ {\bf Work \; done}= \int \vec{\bf F . } \vec{\bf dx} }}}[/tex]
Required Solution :
Since a variable force acts upon the body[tex]\bigstar[/tex] Formula to find the work done
[tex]\leadsto \; {\pink{\boxed{ {\bf Work \; done}= \int\vec{\bf F . } \vec{\bf dx} }}}[/tex]
[tex]\longrightarrow \vec{\sf F} . \vec { \sf dr} = \sf | F | |dx| \cos(\emptyset )[/tex]
[tex]\bigstar[/tex] Here ,
As the force and the displacement are in the same direction the angle between them is 0° [ Cos ( 0 ) = 1 ]
[tex]\longrightarrow \rm Work \; done = \displaystyle \int^4_0 \rm F . dx[/tex]
[tex]\longrightarrow \rm Work \; done = \displaystyle \int^4_0\rm ( 20+ 5x ) \;dx[/tex]
[tex]\longrightarrow \rm Work \; done = \displaystyle \int^4_0 \rm 20 \; dx + \int^4_0 5x \; dx[/tex]
[tex]\longrightarrow \rm Work \; done = \displaystyle 20 \int^4_0 \rm dx + 5 \int^4_0 x \; dx[/tex]
[tex]\bigstar[/tex] Integrating further,
[tex]\longrightarrow \rm Work \; done = 20 \bigg[ x \bigg]^4_0 + 5 \bigg[ \dfrac{x^2}{2} \bigg]^4_0[/tex]
[tex]\longrightarrow \rm Work \; done = 20 \bigg[ 4 - 0 \bigg]+ 5 \bigg[ \dfrac{4^2- 0^2}{2} \bigg][/tex]
[tex]\bigstar[/tex]Calculating the work done,
[tex]\longrightarrow \rm Work \; done = 20 ( 4) + 5 ( 8 - 0 )[/tex]
[tex]\longrightarrow \rm Work \; done = 20 ( 4) + 5 ( 8 )[/tex]
[tex]\longrightarrow {\red{\underline{\underline{\rm{ Work \; done_{ (on \; the \; particle)}= 120 \; Joules }}}}}[/tex]
Therefore :
The work done on the particle is 120 Joules▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
4. arti dari rest your eyes on a couch
istirahatkan matamu di sofa mengistirahatkan mata anda di sofa
5. 3. A block of mass m/ = 4.0 kg is put on top of a block of mass m2 = 5.0 kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12 N must be applied to the top block. The blocks assembly is now placed on a horizontal, frictionless table, as shown in FIGURE 1. Find the magnitudes of a.the maximum horizontal force that can be applied to the lower block so that the blocks will move together b.the resulting acceleration of the blocks.
The maximum horizontal force that can be applied to the lower block is 5 N.The resulting acceleration of the blocks are 21,1 m/s².
Penjelasan dengan langkah-langkah:
Diketahui:
Massa 1 = 4 kg
Berat 1 = 40 N
Massa 2 = 5 kg
Berat 2 = 50 N
F1 = 12 N
Ditanya:
Gaya pada blok paling bawah (F2)?Percepatan (a)?Jawab:
∑Fy1 = 0
N1 - W cos θ = 0
12 - 120 cos θ = 0
12 = 120 c0s θ
cos θ = 0,1
∑Fx1 = m1 x a
120 - T = 4a....(1)
∑Fy2 = 0
N2 - w cos θ = 0
N2 = w cos θ
N2 = 50 (0,1) = 5 N
∑Fx2 = m2 x a
50 sin θ + T = 5a
50 x 1 + T = 5a
50 + T = 5a..........(2)
120 - T = 4a
50 + T = 5a
___________ +
190 = 9a
a = 21,1 m/s²
from that calculation, we know that:
The maximum horizontal force that can be applied to the lower block is 5 N.The resulting acceleration of the blocks are 21,1 m/s².Pelajari lebih lanjut
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6. A 10 kg block with an initial velocity of 10 m/s slides 10 meters across a horizontal surface and comes to a stop. It takes the block 2 seconds to stop. What is the stopping force acting on the block?
Jawaban:
F = 50 N
Penjelasan:
m = 10 kg
v0 = 10 m/s
vt = 0 m/s
Δx = 10 m
t = 2 s
vt = v0 + at
0 = 10 + 2a
-2a = 10
a = -5 m/s²
F = m x a
F = 10 x 5
F = 50 N, with opposite direction
#希望能帮助 (Semoga Membantu:)
7. The graph shows the variation with time of the resultant net force acting on an object. The objecthas a mass of 1 kg and is initially at rest.What is the velocity of the object at a time of 200 ms?A. 8 msB. 16 msC. 8 kmsD. 16 kms
Jawaban:
16 m/s
Penjelasan:
a=F/m=80/1=80 N/kg
v1=v0+at
v1=0 m/s + 80 N/kg*200 ms
v1=16 m/s
8. 1. A block is 25 cm long, 45 cm wide and 10 cm high. then the surface area and the volume of the block are. A block is 25 cm long, 45 cm wide and 10 cm high. then the surface area and the volume of the block areA block is 25 cm long, 45 cm wide and 10 cm high. then the surface area and the volume of the block are
Jawaban:
surface area of the block :
Area = 2 (p×l + p×t + l×t)
= 2 (25×45 + 25×10 + 45×10)
= 2 (1125 + 250 + 450)
= 2 × 1825
= 3650 cm^2
volume of the block :
V = p × l × t
= 25 × 45 × 10
= 11250 cm^3
9. The movement of an object will be ...................... on a smooth surface compared to the movement of an object on a rough surface
Strength ... less on a smooth surface than on a rough surface
10. three strings A, B, and C, are attached to a circular ring, as shown in Figure 4.35. The strings and the ring all lie on a smooth horizontal surface and are at rest. The tension in string A is 8.0 N. Calculate the tension in strings B and C.
Jawaban:
a
maaf kak kalo salah,, semoga bisa bantu
11. The making of cartography relief is based on... A. The extent and depth of earth's surface waters B. The landscape on Earth's surface C. The altitude of a particular place on earth's surface D. The social and economic conditions of a religion
A. The extent and depth of earth's surface waters
12. ...... your oven and ....... it on a flat level surface .
so your oven and inside it on flat level surface(Open) your oven and (put) it on flat level surface
13. Jawab pertanyaan ini dan berikanlah penjelasannya! The spring is placed between the wall and the 10-kg block. If the block is subjected to a force of F = 500 N, determine its velocity when s = 0.5 m. When s = 0, the block is at rest and the spring is uncompressed. The contact surface is smooth. Jawab pertanyaan ini dan berikanlah penjelasannya!
Jawaban:
Jawab pertanyaan ini dan berikanlah penjelasannya! The spring is placed between the wall and the 10-kg
block. If the block is subjected to a force of F = 500 N,
determine its velocity when s = 0.5 m. When s = 0, the
block is at rest and the spring is uncompressed. The contact
surface is smooth. Jawab pertanyaan ini dan berikanlah penjelasannya!
Penjelasan:
Itu jawabannya yang Ngakak ^_^14. Three sleds are being pulled horizontally on frictionless horizontal ice using horizontal ropes. The pull is of magnitude 125 N. Find (a) the acceleration of the system and (b) the tension in ropes A and B.
Jawaban: (a) The acceleration of the system .(b) The Tension in rope A is equal to , and rope B is equal to .
Penjelasan: (a)The expression for acceleration from Newton’s second law is,Here, is net force and is total mass of the system.Substitute for and for.Therefore, the acceleration of the system.
(b)To find the tension in rope A,we consider the system of second and the third boat having mass of.The expression for Tension force from Newton’s second law is,Here, is the mass of the system of the second and third boat and is acceleration.Substitute for and for.Therefore, the tension in rope A is equal to .To find the tension in rope B, we consider the system of only third boat having mass of.The expression for Tension force from Newton’s second law is,Here, is the mass of the system of the third boat and is acceleration.Substitute for and for.Therefore, the tension in rope B is equal to .
15. a car reaches a velocity of 20 m/s with an acceleration of 2 m/s^2 . How far will it travel while it is accelerating if it is (a). initially at rest (b). initially moving at 10 m/s ?
◾ Materi : Kinematika Gerak Lurus
◾ Sub materi : GLBB
◾ Mapel : Fisika
Diketahui :
Vt = 20 m/s
a = 20 m/s²
Ditanya :
s = ?
a) Vo = 0
b) Vo = 10 m/s
Penyelesaian :
*Cari jarak tempuh jika diam
Vt² = Vo² + 2as
20² = 0² + 2(2)s
400 = 4s
s = 100 meter
*Cari jarak tempuh jika Vo = 10 m/s
Vt² = Vo² + 2as
20² = 10² + 2(2)s
400 = 100 + 4s
300 = 4s
s = 75 meter
Jadi, jarak tempuh untuk Vo diam Dan Vo 10 m/s adalah 100 meter dan 75meter
semoga membantu
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-vin
16. FISIKA - KELAS 10 - MOMENTUM DAN IMPULS A 10 kg block is released from rest at point A on track ABCD as shown below. The block collides in a perfectly inelastic manner with another block, initially at rest, of mass 5 kg located just before section BC. The track is smooth except for section BC, of length 6 meters. The blocks hit the spring ( k = 4000 N/m) and compress it a distance of 0.3 meters from its equilibrium position. Determine the coefficient of kinetic friction between the blocks and section BC of the track.
gak bisa ngirim 2 foto, jadi mesti disatuin T-T. kalau gak jelas nanti aku kirim satu satu.
cmiiw.DYNAMICS
• motion and force
• work and energy
m1 = 10 kg
m2 = 5 kg
g = 10 m/s²
h = 10 m
S = 6 m
k = 4000 N/m
∆x = 0,3 m
u = __?
17. A volcanic rock is back, glossy, and smooth. How did this rock most likely form ? A. Rapidly, on the Earth's surface B. Rapidly, beneath the Earth's surface C. Slowly, on the Earth's surface D. Slowly, beneath the Earth's surface
Jawaban:
pppppppppppppp
Penjelasan:
iqidhshqhdjs
18. A 5g bullet moving at 100 m/s collides a wooden target, initially at rest. The mass of wooden target is 1kg. If after the collision the bullet is embed in the wooden block, what is the final speed of the wooden block?
This is innelastic collision. so we can use formula:
m1.v1+m2v2=(m1+m2)vf
5x100+1kgx0=(5+1000)vf
500+0=1005Vf
Vf=500/1005=0.49 m/s
19. Force of . . . is less on a smooth surface than on a rough surface.
Jawaban :
Kekuatan ... lebih sedikit pada permukaan yang halus dibandingkan pada permukaan yang kasar
20. Push a car across a frictionless surface,you apply a 40N force for a period of 4 seconds.the car goes from rest to velocity of 5m/s. What is the mass of car?
Calculate the acceleration
Vf = Vi + a.t
5 = 0 + a.4
4a = 5
a = 5/4 = 1,25 m/s²
Find the mass
F = m.a
m = F/a = 40/1,25
m = 32 kg
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