A mass of 5.2 kg lies at the top of a frictionless incline. It is inclined at an angle such that the normal force on the mass is 29 Newtons. If the length of the incline is 78 meters and the mass is originally at rest at the top of the incline and then released, how long does it take for the mass to reach the bottom of the incline in seconds?

To find out how long it takes for the mass to reach the bottom of the incline, we can use the concept of mechanical energy conservation. The initial potential energy (PE) of the mass at the top of the incline will be converted into kinetic energy (KE) when it reaches the bottom.

The potential energy at the top is given by the equation PE = m * g * h, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the vertical height of the incline. In this case, the vertical height is given by h = l * sin(θ), where l is the length of the incline and θ is the angle of the incline.

So, the potential energy at the top is PE = 5.2 kg * 9.8 m/s² * (78 m * sin(θ)).

The kinetic energy at the bottom is given by the equation KE = 0.5 * m * v², where v is the velocity of the mass at the bottom.

Since mechanical energy is conserved, the potential energy at the top is equal to the kinetic energy at the bottom, so we have:

PE = KE
5.2 kg * 9.8 m/s² * (78 m * sin(θ)) = 0.5 * 5.2 kg * v²

We can cancel out the mass:

9.8 m/s² * (78 m * sin(θ)) = 0.5 * v²

Now, we solve for v:

v² = 2 * 9.8 m/s² * 78 m * sin(θ)
v² = 1525.04 * sin(θ)

Taking the square root of both sides, we get:

v = √(1525.04 * sin(θ))

Now, to find the time it takes for the mass to reach the bottom of the incline, we can use the equation:

v = d/t

where d is the length of the incline and t is the time taken.

Rearranging the equation, we have:

t = d/v

Substituting the values, we have:

t = 78 m / √(1525.04 * sin(θ))

Now, you can plug in the value of the angle (θ) to calculate the time it takes for the mass to reach the bottom of the incline in seconds.