How far will 20 N of force stretch a spring with a spring constant of 140 N/m?
1 answer:
Answer:
0.143 m
Explanation:
The relationship between force applied on a string and stretching of the spring is given by Hooke's law:
where
F is the force exerted on the spring
k is the spring constant of the spring
x is the stretching of the spring from its equilibrium position
In this problem, we have:
F = 20 N is the force applied on the spring
k = 140 N/m is the spring constant
Solving for x, we find how far the spring will stretch:
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Answer:
a) Initial speed of the ball = 14.45 m/s
b) At height 6 m speed of ball = 9.55 m/s
c) Maximum height reached = 10.65 m
Explanation:
a) We have equation of motion , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.
s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8
Substituting
Initial speed of the ball = 14.45 m/s
b) We have equation of motion , where v is the final velocity
s = 6 m, u = 14.45 m/s, a = -9.8
Substituting
So at height 6 m speed of ball = 9.55 m/s
c) We have equation of motion , where v is the final velocity
u = 14.45 m/s, v =0 , a = -9.8
Substituting
Maximum height reached = 10.65 m