mr. slott at his fighting weight of 1000N jumps down from a platform meters above a trampoline with a spring constant (k) of 500N/M. How much does Mr.Slott strech the trampoline (x) before it brings him to a stop
d = 1000N * 1m/500N = 2 m.
A/c to definition of weight
F=W
A/c to hooke's law:
F=kx
Now, if F=W then;
W=kx
1000=500. x
X=1000÷500
X=2m
To determine how much the trampoline stretches (x) when Mr. Slott jumps on it, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be represented as:
F = -kx
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement of the spring.
In this case, the force exerted by the spring (F) is equal to Mr. Slott's weight, which is 1000N, since he is at his fighting weight.
Therefore, the equation becomes:
1000N = -kx
Given that the spring constant (k) is 500N/m, we can substitute it into the equation:
1000N = -500N/m * x
To solve for x, we rearrange the equation:
x = -1000N / (-500N/m)
x = 2 meters
Therefore, Mr. Slott stretches the trampoline by 2 meters before it brings him to a stop.