The work to stretch a spring from its rest position is
(1/2) (spring constant) (distance of the stretch)²
E = 1/2 k x² .
You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write
1700 joules = 1/2 k (3m)²
1 joule = 1 newton-meter
1700 N-m = 1/2 k (3m)²
Multiply each side by 2: 3400 N-m = k · 9m²
Divide each side by 9m² k = 3400 N-m / 9m²
= (377 and 7/9) newton per meter
Answer:
Woke done, W = 4156.92 Joules
Explanation:
The work done by the force can be calculated as :


is the angle between force and the displacement
It is assumed to find the work done for the given parameters i.e.
Force, F = 30 N
Distance travelled, s = 160 m
Angle between force and displacement, 
Work done is given by :


W = 4156.92 Joules
So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.
Answer:the 90% is spent performing other functions
Explanation:
<span>C. Mao Zedong
Hope this helps!~</span>
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s