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Sagot :
Answer:
distance travelled by the block is 0.796 m
{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }
Explanation:
Given that;
mass of block m = 0.200 kg
distance travelled d = 0.796 m
time t = 2.00 s
m₂ = 0.400 kg
If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?
Now, using the second equation of motion;
d = ut + ([tex]\frac{1}{2}[/tex] × at²)
as the object started from rest, u=0
so, we substitute
0.796 = 0×2 + ([tex]\frac{1}{2}[/tex] × a(2)²)
0.796 = 0 + ([tex]\frac{1}{2}[/tex] × 4a)
0.796 = 2a
a = 0.796 / 2
a = 0.398 m/s²
using first equation of motion
[tex]V_{f}[/tex] = u + at
we substitute
[tex]V_{f}[/tex] = 0 + 0.398 × 2
[tex]V_{f}[/tex] = 0.796 m/s
now, average velocity is given as;
[tex]V_{avg}[/tex] = ( 0.796 m/s + 0 ) / 2
[tex]V_{avg}[/tex] = ( 0.796 m/s + 0 ) / 2
now, distance as the block moves in 2s will be;
D = [( 0.796 m/s + 0 ) / 2 ] × 2
D = 0.796 m
Therefore, distance travelled by the block is 0.796 m
{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }
The distance traveled by the object with the uniform motion can be given by the second equation of the motion.
The distance traveled by the big block with weight in 2 seconds is.
What is second equation of motion?
The distance traveled by the object with the uniform motion can be given by the second equation of the motion. It can be given as,
[tex]s=ut+\dfrac{1}{2} at^2[/tex]
Given information-
The mass of the small block is 0.200 kg.
The total distance traveled by the block is 0.796.
Initial velocity of small block is zero.
Total time taken by the block to travel this distance is 2 seconds.
Put the values in the above equation as,
[tex]0.796=0\times 2+\dfrac{1}{2} a\times 2^2\\0.796=2a\\a=0.398[/tex]
Thus the acceleration of the small block is 0.398 meter per second.
Now the mass is doubled which is, 0.400 kg. As the acceleration does not depends on the mass, thus the acceleration for both cases is 0.398.
The velocity of the big block can be given as,
[tex]V=u+0.398\times2\\V=0+0.796\\V=0.796[/tex]
The velocity of the big block is 0.796.
The average velocity of the big block is given by,
[tex]V_{avg}=(\dfrac{0.796+0}{2} })\\V_{avg}=0.398[/tex]
The distance traveled by the object is the ratio of the velocity of the body to the time taken by it. Thus the distance traveled by the big block in 2 seconds is,
[tex]d={{0.398} \times2}\\d=0.796[/tex]
Thus the distance traveled by the big block with weight in 2 seconds is.
Learn more about the second equation of motion here;
https://brainly.com/question/8898885
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