Let us use the formula for Newton's Second Law of Motion:
Net force = Mass*Acceleration
Net force = Applied Force - μ*Normal Force
where μ is the coefficient of kinetic friction
Normal Force = Force due to gravity = mass*gravity
Normal Force = (210 kg)(9.81 m/s²) =<em> 2,060.1 N</em>
Then,
Net force = 4100 - 0.38*2060.1 = 3317.162 N
3317.162 N = (210 kg)(a)
Solving for acceleration,
<em>a = 15.796 m/s²</em>
The Atwood's machine is in motion starting from rest, then Vf = Vo + a(t).
<span>Final Velocity is given as 6.7 m/s and the time is 1.9 s thus 6.7= 0+ a(1.9) </span>
<span>then a = 6.7/1.9 = 3.526 m/s². </span>
<span>The Atwood's Machine also has the formula d= distance = 1/2a(t²) </span>
<span>distance given is 6.365 m , then 6.365 = 1/2 a (1.9)², </span>
<span>a = 3.526 m/s² the same acceleration. </span>
<span>a= g(m1-m2) / m1+m2) </span>
<span>m1a + m2a = m1g - m2g </span>
<span>m1a - m1g = -m2g - m2a </span>
<span>3.526 m1 - 9.81 m1 = -9.81m2 - 3.526 m2 </span>
<span>-6.28 m1 = -13.34 m2 </span>
<span>0.47 m1= m2 </span>
<span>if 24J = 1/2mv² </span>
<span>then 24J = 1/2 m1 ( 6.7)² </span>
<span>48/ 44.89 = m1 </span>
<span>1.069 kg = m1 , then </span>
<span>0.47(1.069) = m2 </span>
<span>0.503 kg = m2</span>
At the beginning of a basketball game, a referee tosses the ball straight
up with a speed of 4.6 m/s. a player cannot touch the ball until after it reaches its maximum height and begins to fall down. what is the minimum time that a player must wait before touching the ball? s
-----------------------
kinemaic equation
v=u-at
0=4.6-9.81xt
t=4.6/9.81 ... about half a second
Answer:
Forces Acting in the Same Direction
If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.
Explanation:
They will become identical as if they were never broken
