Answer:
b. It is dropped
Explanation:
If the initial velocity is zero, the object move from rest. That happens if the object is dropped
Answer:
0.6983 m/s
Explanation:
k = spring constant of the spring = 0.4 N/m
L₀ = Initial length = 11 cm = 0.11 m
L = Final length = 27 cm = 0.27 m
x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m
m = mass of the mass attached = 0.021 kg
v = speed of the mass
Using conservation of energy
Kinetic energy of mass = Spring potential energy
(0.5) m v² = (0.5) k x²
m v² = k x²
(0.021) v² = (0.4) (0.16)²
v = 0.6983 m/s
The average power is 
Explanation:
First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:
where
:
is the final kinetic energy of the car, with
m = 2000 kg is the mass of the car
v = 60 m/s is the final speed of the car
is the initial kinetic energy of the car, with
u = 30 m/s is initial speed of the car
Soolving:
Now we can find the power required for the acceleration, which is given by
where
t = 9 s is the time elapsed
Solving:
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Answer:
10.6cm
Explanation:
We are given 5.3cm below the starting point (spring extension).
Therefore, to find static vertical equilibrium, we use the equation:
kx = mg
Where:
k = spring constant =
=mg/5.3 kg/s²
We are told the object was dropped from rest.
Therefore:
loss in potential energy = gain in spring p.e
Let's use the expression:
mgx = ½kx²
We are asked to find the stretch at maximum elongation x.
To find x, we make x subject of the formula.
Therefore, we have:
x = 2mg/k (after rearranging the equation above)
x = (2mg) / (mg/5.3)
x = 10.6cm
