In a movie, a stuntman places himself on the front of a truck as the truck accelerates. The coefficient of friction between the stuntman and the truck is \(0.65 .\) The stuntman is not standing on anything but can "stick" to the front of the truck as long as the truck continues to accelerate. What minimum forward acceleration will keep the stuntman on the front of the truck?

We can start by identifying the forces acting on the stuntman. There are two main forces: 1. Gravitational force (Fg) acting downwards. 2. Frictional force (Ff) acting upwards along the front of the truck, which opposes the gravitational force.

According to Newton's second law of motion, the net force acting on an object is equal to the product of the object's mass (m) and acceleration (a). In our case, the net force acting on the stuntman is the difference between the frictional force and the gravitational force: Net force (F) = Ff - Fg

To find the frictional force acting on the stuntman, we'll use the equation: Ff = μ * Fn Where μ is the coefficient of friction (0.65) and Fn is the normal force acting on the stuntman, which is equal to Fg (m * g) since there is no vertical acceleration.

Now, we can rewrite the net force equation as: F = μ * Fg - Fg

We know that F = m * a, and we want to find the acceleration (a). We can rewrite the equation in terms of acceleration: a = F / m Substitute the net force equation from Step 4: a = (μ * Fg - Fg) / m Since Fg = m * g, we can replace Fg with m * g: a = (μ * m * g - m * g) / m Cancel out the mass (m) from the equation: a = (μ * g - g) Now, plug in the values for μ (0.65) and g (9.81 m/s²): a = (0.65 * 9.81 - 9.81) a = 1.635 - 9.81 a ≈ -8.175 m/s² However, the negative sign indicates that the stuntman is actually accelerating backward relative to the front of the truck. He's staying in place relative to the ground, but the truck is moving forward. So, we can take the absolute value of the acceleration: a = 8.175 m/s² (approx.)

The minimum forward acceleration required to keep the stuntman on the front of the truck is approximately 8.175 m/s².