Source: Principles of Physics
Problem
When a click beetle is upside down on its back, it jumps upward by suddenly arching its back. Videotape of a certain click-beetle jump shows that a beetle of mass $m = 4.0 \times 10^{-6}$ kg moved directly upward by $d = 0.77$ mm during the launch and then to a maximum height of $h = 0.30$ m.
- During the launch, what is the average magnitude of the external force on the beetle's back from the floor?
- During the launch, what is the average magnitude of the acceleration of the beetle in terms of $g$?
[Q1] $F_{avg} = 1.5 \times 10^{-2}$ N [Q2] $a = 390g$
The beetle's motion consists of two phases: the launch (accelerating upwards over distance $d$) and the free-fall (decelerating upwards to height $h$).
First, analyze the free-fall phase to find the launch velocity $v_L$. The beetle starts with velocity $v_L$ and reaches $v_f = 0$ at height $h$. Using the kinematic equation $v_f^2 = v_i^2 + 2a_y \Delta y$:
$$0^2 = v_L^2 + 2(-g)h$$ $$v_L^2 = 2gh$$Next, analyze the launch phase to find the average acceleration $a$. The beetle starts from rest ($v_0 = 0$) and reaches velocity $v_L$ over a distance $d$. Using the same kinematic equation:
$$v_L^2 = 0^2 + 2ad$$ $$a = \frac{v_L^2}{2d}$$Substitute the expression for $v_L^2$:
$$a = \frac{2gh}{2d} = \frac{gh}{d}$$The acceleration in terms of $g$ is $a/g = h/d$. This solves [Q2].
To find the average external force from the floor, $F_{avg}$, apply Newton's second law during the launch phase. The net force on the beetle is $F_{net} = F_{avg} - mg$.
$$F_{net} = ma$$ $$F_{avg} - mg = ma$$ $$F_{avg} = m(a+g)$$Substitute the expression for $a$:
$$F_{avg} = m\left(\frac{gh}{d} + g\right) = mg\left(\frac{h}{d} + 1\right)$$This solves [Q1].