Question

The cart shown in fig.4-55 lies on a plane tilted at an angle of 22 degrees o the horizonal with .12 coefficient for kinetic friction. Determine the acceleration ofthe carton as it slides down the plane. If the carton starts from rest 9.3 meters up the plane from it's base, what will be the cartons speed when it reaches thebottom.?

Answer 1

You did not include the figure, so I am guessing somewhat about the layout, especially for the second part. Do you mean 9.3 meters of slidingbefore hitting the bottom, in other words a 9.3 meter hypotenuse for the 22 degree right triangle?

The acceleration is calculated like this: the cart weighs mg; the force pulling it down the plane is mg sin θ where θ is 22 degrees; the normalforce is mg cos θ so the frictional force, opposing the slide down, is μ mg cos θ. Subtracting that from the first force, we have:

F_net = mg(sin θ - μ cosθ) and since F = ma, we have

a_net = g(sin θ - μ cos θ) = (9.81 m/s²)(sin 22° - .12 cos 22°)

= (9.81 m/s²)(0.374607 - 0.12 * 0.927184)

= 2.58 m/s² which is the answer to the first part.

Assuming it slides 9.3 meters down the plane, from basic kinematics we have

v = (2ad)^1/2 = (2 * 2.58 m/s² * 9.3 m)^1/2 = 6.93 m/s