A 3.75 kg cat leaps off the floor with a launch speed of 4.0 m/s at 30° to the ground to pounce on a mouse which is running away from it at 1.5 m/s. The launch lasts 0.15 s, after which the cat is airborne.

3.1
Determine the net average force on the cat as it leaps.

Go ask Gregor

It's all good to say that Brandon but the fact is you googled as well and came to this link. Hahahahaha

To determine the net average force on the cat as it leaps, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

First, we need to find the acceleration of the cat during the launch phase. The launch is essentially a projectile motion problem, where the cat is moving at an angle of 30° with respect to the ground.

The horizontal component of the cat's velocity, v_x, can be found using the launch speed and the angle: v_x = v * cos(θ) = 4.0 m/s * cos(30°) = 3.46 m/s.

The vertical component of the cat's velocity, v_y, can also be found using the launch speed and the angle: v_y = v * sin(θ) = 4.0 m/s * sin(30°) = 2.00 m/s.

The time of flight t during the launch phase can be calculated using the vertical component of velocity: v_y = g * t, where g is the acceleration due to gravity (9.8 m/s^2). Solving for t, we have: t = v_y / g = 2.00 m/s / 9.8 m/s^2 = 0.204 s.

Since we are given that the launch lasts for 0.15 s, it means that the cat is airborne for only 0.15 s - 0.204 s = -0.054 s. However, this negative value does not make sense, so we conclude that the cat does not complete its leap and falls to the floor.

Since the cat falls back to the floor, the net average force on the cat during the leap can be calculated using only the vertical component of velocity, as follows:

Using the equation for average acceleration (\(a_{avg} = \frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\) and \(\Delta t\) is the duration), we have that:

\(a_{avg} = \frac{v_f - v_i}{\Delta t} = \frac{0 - v_y}{0.15 s}\).

Since the initial vertical velocity \(v_y\) is 2.00 m/s, the average acceleration becomes:

\(a_{avg} = \frac{0 - 2.00 m/s}{0.15 s} = -13.3 m/s^2\).

Finally, we can multiply the cat's mass (3.75 kg) with the average acceleration to find the net average force:

\(F_{avg} = m * a_{avg} = 3.75 kg * (-13.3 m/s^2) = -49.8 N\).

Therefore, the net average force on the cat as it leaps is approximately -49.8 Newtons downward.