A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,
1 answer:
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Answer:
a) v = 7,207 m / s
, b) a = 42.3 m / s²
Explanation:
We will solve this exercise using the concept of mechanical energy, We will write it in two points before the car touches the springs and in point of maximum compression
Initial
Em₀ = K = ½ m v²
Final
= 2 Ke = ½ k x²
The two is placed because each barred has two springs and each does not exert the same force
Emo =
½ m v² = 2 ½ k x²
v = √(2k/m) x
v = √ (2 3,134 10⁵/4550) 0.614
v = 7,207 m / s
Let's take this speed to km / h
v = 5,096 m / s (1km / 1000m) (3600s / 1h)
v = 25.9 km / h
This speed is common in school zones
Let's use kinematics to calculate the average acceleration
vf² = v₀² - 2 a x
0 = v₀² - 2 a x
a = v₀² / 2 x
a = 7,207²/2 0.614
a = 42.3 m / s²
We buy this acceleration with the acceleration of gravity
a / g = 42.3 / 9.8
a / g = 4.3
This acceleration is well below the maximum allowed
Answer:
v = at + u
Explanation:
acceleration, a = constant
As we know that acceleration is the rate of change of velocity
integrate on both sides
v = at + u
Where, u is the integrating constant and here it is equal to the initial velocity
Now we know that the rate of change of displacement is called velocity
Integrate on both sides
where, xo is the integrating constant which is initial position of the particle.