Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl
1 answer:
Answer:
t = 1.41 sec.
Explanation:
If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.
First, we need to find the value of acceleration, which is the same for both blocks.
If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:
F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)
⇒ a = ( = g/5 m/s²
Once we got the value of a, we can use for instance this kinematic equation, and solve for t:
Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.
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The time needed for the 7th car to pass is 13.2 s
Explanation:
The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.
We start by analzying the motion of the first car, by using the equation:
where
s is the distance covered by the first car in a time t, which corresponds to the length of one car
u = 0 is the initial velocity
a is the acceleration
t = 5.0 s is the time
The equation can be rewritten as
where L is the length of one car.
The same equation can be written considering the first 7 cars:
where
7L is the distance covered by the 7 cars
t' is the time needed
We still have
u = 0
And the acceleration is constant so it is
Substituting into the equation, we can find t':
In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.
Learn more about accelerated motion:
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