Question

An insect weighing 5 g takes off vertically for a distance of 100 cm with a speed of 4 m s\(^{-1}\) by using its hind legs. Ignoring the resistance due to air, the magnitude of the average net force exerted by the hind legs during the take-off is ............ N. (rounded off to 3 decimals)

Hint:

To calculate the force exerted during take-off, use Newton's second law to find the net force needed to accelerate the insect, and add the force required to overcome gravity.

Answer 1

Step 1: Given values.
- Mass of the insect, \( m = 5 \, \text{g} = 0.005 \, \text{kg} \)
- Distance traveled during take-off, \( d = 100 \, \text{cm} = 1 \, \text{m} \)
- Initial speed, \( u = 0 \, \text{m/s} \) (since the insect starts from rest)
- Final speed, \( v = 4 \, \text{m/s} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)
Step 2: Use of kinematic equation to find acceleration.
To find the acceleration during take-off, we use the following kinematic equation: \[ v^2 = u^2 + 2ad \] Substituting the known values: \[ (4)^2 = 0^2 + 2 \times a \times 1 \] \[ 16 = 2a \quad \Rightarrow \quad a = 8 \, \text{m/s}^2 \] Step 3: Using Newton's second law to find the net force.
The net force required for the insect to take off is the sum of the force needed to overcome gravity and the force needed to accelerate. The total force \( F_{\text{net}} \) is given by: \[ F_{\text{net}} = ma \] Substituting the mass and acceleration: \[ F_{\text{net}} = 0.005 \times 8 = 0.04 \, \text{N} \] Additionally, the force required to overcome gravity is: \[ F_{\text{gravity}} = mg = 0.005 \times 9.8 = 0.049 \, \text{N} \] So, the total net force exerted by the hind legs during take-off is: \[ F_{\text{legs}} = F_{\text{net}} + F_{\text{gravity}} = 0.04 + 0.049 = 0.089 \, \text{N} \] Final Answer: \[ \boxed{0.089 \, \text{N}} \]