The maximum torque output from the engine of a new experimental car of mass m is τ. The maximum rotational speed of the engine is ω. The engine is designed to provide a constant power output P. The engine is connected to the wheels via a perfect transmission that can smoothly trade torque for speed with no power loss. The wheels have a radius R, and the coefficient of static friction between the wheels and the road is µ.

What is the maximum sustained speed v the car can drive up a 30 degree incline? Assume no frictional losses and assume µ is large enough so that the tires do not slip.

Normal force = m g cos 30

max friction force = mu m g cos 30

power in = power out
P = (mu m g cos 30) v

It says that mu is large enough that the tires don't slip

the answer is v = 2P/mg
of why there is not root(3) in the denominator i am not sure but that is what the USAPho test solution says.

To find the maximum sustained speed of the car up a 30-degree incline, we need to consider the forces acting on the car.

First, let's calculate the gravitational force acting on the car. The gravitational force formula is given by:

F_gravity = m * g

where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Next, let's calculate the force required to overcome the incline. The force required to overcome the incline is the sum of the gravitational force component parallel to the incline and the force required to counteract the friction.

The gravitational force component parallel to the incline is given by:

F_parallel = F_gravity * sin(θ)

where θ is the angle of the incline (30 degrees in this case).

The force required to counteract the friction is:

F_friction = F_gravity * cos(θ) * µ

where µ is the coefficient of static friction between the wheels and the road.

Now, let's calculate the total force required to climb the incline:

F_total = F_parallel + F_friction

We know that power, P, is the rate at which work is done or force is applied over a distance:

P = F_total * v

where v is the speed of the car. We can rearrange this equation to solve for v:

v = P / F_total

Substituting the values we have, we can calculate the maximum sustained speed of the car up the 30-degree incline.

To find the maximum sustained speed of the car driving up a 30-degree incline, we need to consider the balance of forces acting on the car.

The force of gravity acting on the car can be split into two components: one parallel to the incline and the other perpendicular to the incline. The component of gravity parallel to the incline is given by mg * sin(30), where m is the mass of the car and g is the acceleration due to gravity.

The force required to overcome the incline is the sum of the force of gravity parallel to the incline and the force of friction. The force of friction can be calculated using the coefficient of static friction µ and the normal force, which is mg * cos(30), where m is the mass of the car and g is the acceleration due to gravity.

The power output P of the engine is given by the torque τ multiplied by the rotational speed ω. Since the engine provides a constant power output, we have P = τ * ω.

The force required to overcome the incline is equal to the power output divided by the velocity v. So we have:

mg * sin(30) + µ * mg * cos(30) = P / v

Rearranging the equation, we can solve for v:

v = P / (mg * sin(30) + µ * mg * cos(30))

Substituting the expression for P (P = τ * ω), we get:

v = (τ * ω) / (mg * sin(30) + µ * mg * cos(30))

So, the maximum sustained speed v the car can drive up a 30-degree incline is given by this equation. To find the value of v, you will need to know the values of τ (maximum torque output from the engine), ω (maximum rotational speed of the engine), m (mass of the car), R (radius of the wheels), µ (coefficient of static friction between the wheels and the road), and g (acceleration due to gravity).