A loaded spring launches a 2.50 kg block, using a force of 450 N. If the change in
1 answer:
Answer:
37.5 seconds
Explanation:
The given parameters are;
The mass of the block on the spring, m = 2.50 kg
The force with which the loaded spring launches the block, F = 450 N
The change in momentum of the block, Δp = 12.0 kg·m/s
We have;
Let the force with which the block was launched = The net force,
By Newton's second law of motion, we have;
F = = Δp × Δt
Where;
Δt = The time the block is in contact with the spring
Therefore;
By plugging in the values for and Δp, we have;
The time duration the block is in contact with the spring, Δt = 37.5 seconds.
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Answer:
a) v = 16.57 m / s, b) a = 19.6 m / s², d) N = 1.76 10³ N, N / W = 3
Explanation:
This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.
Let's answer the questions.
a) For this part we can use energy considerations.
Starting point. The upper part of the trajectory indicates that the arm is horizontally
Em₀ = U = m g h
in this case h = r
Final point. For lower of the trajectory
Em_f = K = ½ m v²
as they indicate that there is no friction
Em₀ = em_f
mgh = ½ m v²
v =
let's calculate
v =
v = 16.57 m / s
b) the centripetal acceleration has the formula
a = v² / r
a = 16.57² / 14.0
a = 19.6 m / s²
c) see attached where the diagram is
where N is the normal and w the weight
d) let's use Newton's second law
N-W = m a
N - mg = m ar
N = m (g + a)
let's calculate
N = 60.0 (9.8 + 19.6)
N = 1.76 10³ N
the relationship with weight is
N / W = 1.76 10³/( 60 9.8)
N / W = 3
normal is three times greater than body weight
e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it
