Answer: An object's velocity (a vector) does not change if and only if the net force acting on the object is zero. In other words, if there is no net force on an object, its speed and direction of motion do not change (including if it is at rest).
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assuming the reference line to measure the height for gravitational potential energy lying at the equilibrium position
m = mass attached to the spring = 10.00 kg
k = spring constant of the spring = 250 N/m
h = height of the mass above the reference line or equilibrium position = 0.50 m
x = compression of the spring = 0.50 m
v = speed of mass = 2.4 m/s
A = maximum amplitude of the oscillation
v' = speed of mass at the maximum amplitude location = 0 m/s
using conservation of energy between the point where the speed is 2.4 m/s and the highest point at which displacement is maximum from equilibrium
kinetic energy + spring potential energy + gravitational potential energy = kinetic energy at maximum amplitude + spring potential energy at maximum amplitude + gravitational potential energy at maximum amplitude
(0.5) m v² + m g h + (0.5) k x² = (0.5) m v'² + m g A + (0.5) k A²
inserting the values
(0.5) (10) (2.4)² + (10) (9.8) (0.50) + (0.5) (250) (0.50)² = (0.5) (10) (0)² + (10) (9.8) A + (0.5) (250) A²
109.05 = (98) A + (125) A²
A = 0.62 m
45° × π/180 = 0,7854 rad
ω = θ /t
ω = 0.7854 / 0.75 = 1.0472 rad/s
Answer:
5 m
35 m
Explanation:
From the formula
s = ut + 1 /2 g t²
u = 0 ; t = 1.
s = 1 / 2 x 10 x 1 = 5 m
In first second distance traveled is 5 m .
Time to travel 80 m is t
80 = 0 + .5 x 10 t²
t = 4s
In last one second
s = u +
= 0 + (2 x4 -1 )x 10 /2 = 35 m .