Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.
1 answer:
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Answer:
a = 3.61[m/s^2]
Explanation:
To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.
In this case we have:
Refer to the diagram shown below.
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).