Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
40 N/m
Explanation:
F = -kx (This is the Hooke's Law equation)
F is the force the spring exerts = 8 N
-k = spring constant
x = displacement (The distance stretched past it's natural length) = 20cm
x needs to be in meters, and 20 cm is = to 0.2 meters
Finally:
8N = -k (0.2m)
-k = 8N / 0.2 m
k = -40 N/m
The answer for this question should be TRUE
