Question

An air cart of mass m is slid up and allowed to slide back down a frictionless air track. Initially, the cart is pushed to get it moving, but the motion we are analyzing is after the initial pushing force has ended. What velocity will the cart have at the bottom of the incline (the same place the force ended) if the cart slides up a 10° incline a distance of 1.5 meters before stopping and sliding back down?

Answer 1

The velocity of the cart at the bottom of the incline is approximately 1.34 m/s.

Step-by-step explanation:

To find the velocity of the cart at the bottom of the incline, we can use the principle of conservation of mechanical energy. The potential energy of the cart at the top of the incline is converted into kinetic energy at the bottom. Since the incline is frictionless, there is no work done by friction. Therefore, the mechanical energy is conserved.

The potential energy at the top is given by PE = mgh, where m is the mass of the cart, g is the acceleration due to gravity, and h is the vertical height of the incline.

The kinetic energy at the bottom is given by KE = 0.5mv^2, where v is the velocity of the cart at the bottom.

Setting the initial potential energy equal to the final kinetic energy:

mgh = 0.5mv^2

Canceling out the mass and solving for v:

v = sqrt(2gh)

Substituting the given values of g = 9.8 m/s^2 and the height h = 1.5sin(10°) m:

v = sqrt(2 * 9.8 * 1.5sin(10°))

Calculating the value of v gives:

v ≈ 1.34 m/s