Explanation:
We have,
Mass of an object is 0.5 kg
Force constant of the spring is 157 N/m
The object is released from rest when the spring is compressed 0.19 m.
(A) The force acting on the object is given by :
F = kx

(B) The force is simply given by :
F = ma
a is acceleration at that instant

Answer:
The duration of the impact is 0.005384 seconds
Explanation:
Given
m = 0.43 kg
v = 5.2 m/s
x = 0.014 m
Knowing the formulas

Answer:
2 m/s²
Explanation:
the equations of motion are
S= ut +½at²
v² = u²+ 2as
v = u + at
s = (u+v)/2 × t
From the parameters given
u = 0m/s this is because it starts from rest
Distance (s) = 9m
Time (t) = 3s
Based on this the first equation would be used
s = ut + ½at²
Input values
9 = 0×3 + ½ × a x 3²
9 = 0 + 9a/2
9 = 4.5a
Divide both sides by 4.5
a = 9 / 4.5 m/s²
a = 2 m/s²
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Answer:
<h2>1567.09 N/m</h2>
Explanation:
Step one:
given data
mass m=5kg
compression x= 3.13cm to m= 0.0313m
<em>According to Hooke's law, provided the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force</em>
F=ke
where
k= spring constant in N/m
e= extension/compression in
Step two:
assume g= 9.81m/s^2
F=mg
F=5*9.81
F=49.05N
substitute in the expression F=ke
49.05=k*0.0313
k=49.05/0.0313
k=1567.09 N/m
<u>The force constant (in N/m) of the spring is 1567.09 N/m</u>