A 42kg girl jumps on a trampoline, after stretching to its bottom limit; the trampoline exerts an average upward force on the girl over a diplacement of 0.50m. During the time that the trampoline is pushing her up, se experiences an average acceleration of 65.0m/s^2. Her velocity at the moment that she leaves the trampoline is 9.4m/s[up]
a) What is the average force that the trampoline exerts on the girl?
b) How much high does she bounce?
To answer these questions, we can use the laws of physics, specifically Newton's second law and the laws of motion.
a) To find the average force exerted by the trampoline on the girl, we can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a). The equation is F = m * a.
Given:
- Mass of the girl (m) = 42 kg
- Acceleration (a) = 65.0 m/s^2 (upward, as specified in the problem)
Plugging in the values, we can calculate the force:
F = 42 kg * 65.0 m/s^2 = 2730 N
Therefore, the average force exerted by the trampoline on the girl is 2730 Newtons.
b) To find the height the girl bounces to, we can use the kinematic equation that relates displacement, initial velocity, final velocity, and acceleration. The equation is:
v^2 = u^2 + 2as
Where:
- v = final velocity (9.4 m/s [up], as specified in the problem)
- u = initial velocity (which is zero when she hits the trampoline)
- a = acceleration (65.0 m/s^2 [up], as specified in the problem)
- s = displacement (unknown, what we want to find)
To solve for s, we rearrange the equation:
s = (v^2 - u^2) / (2a)
Plugging in the values:
s = (9.4)^2 / (2 * 65.0) = 0.67846 m
Therefore, the girl bounces to a height of approximately 0.68 meters.
avgforce*distance=1/2 mass*velocity^2
solve for avgforce.
How high?
1/2 mass*velocity^2=mass*gh
or 1/2 velocity^2/g= height