An object with a mass of 10 kg is initially at rest at the...

An object with a mass of 10 kg is initially at rest at the top of a frictionless inclined plane that rises at 30° above the horizontal. At the top, the object is initially 8.0 m from the bottom of the incline, as shown in the figure. When the object is released from this position, it eventually stops at a distance d from the bottom of the inclined plane along a horizontal surface, as shown. The coefficient of kinetic friction on the hill is 0.30 and coefficient of kinetic friction between the horizontal surface and the object is 0.20, and air resistance is negligible. Find the distance d. (take g = 10.0 m/s²) 8.0 m d 30° - 69.6 m - 9.6 m - 8.0 m - 14.0 m - 4.0 m

Transcript

00:01 So in this question we have an incline at an incline of theta equals 30 degrees.

00:10 And we have a block that starts off at a distance of 8 metres.

00:17 So l equals 8 metres up the hill.

00:23 And the mass of the object is 10 kilograms.

00:26 M equals 10 kilograms.

00:30 And it's initially at rest.

00:31 We're going to take g is 10 meters per year.

00:34 Second squared.

00:36 The coefficient of kinetic friction on the hill is mu equals 0 .30 and on the flat surface it's mu equals 0 .20 and the object will slide down the hill and then slide along the flat bit for a distance d and it stops here.

00:59 So how are we going to work out what d is? well, what's going to be the work done by friction? the work done by friction is going to be the force of friction on the incline times the distance travelled on the incline, which is l, plus the force of friction on the flat surface times deep.

01:25 But the work done by friction is going to be equal to the work done by gravity, because gravity will speed the box up, and friction will slow it down.

01:34 And the work done by gravity is the mass of the box times g times the height that it starts off with, which is l -sign theta.

01:47 Okay, so now we need to work out what these forces of friction are.

01:52 Let's work this out.

01:53 So on the incline, our box has a mass of 10 kilograms, so we have m -g acting downwards, and the normal component of that is going to be m -g -cos -theta, which means we'll have a normal force, n equals m -g -g -cos -theta, and a friction force f on the incline is mu on the incline mg cos theta.

02:21 And this mu on the incline we know is 0 .3...

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