A 4.8 -kg block on a 39^∘ inclined plane is acted on by a horizontal force of 46 N (see Fig. 5-

A 4.8 -kg block on a 39^∘ inclined plane is acted on by a...

A $4.8$ -kg block on a $39^{\circ}$ inclined plane is acted on by a horizontal force of $46 \mathrm{~N}$ (see Fig. $5-35$ ). The coefficient of kinetic friction between block and plane is $0.33 .$ (a) What is the acceleration of the block if it is moving up the plane? $(b)$ With the horizontal force still acting, how far up the plane will the block go if it has an initial upward speed of $4.3 \mathrm{~m} / \mathrm{s}$ ? (c) What happens to the block after it reaches the highest point?

Key Concepts

Kinematics of Uniformly Accelerated Motion

The equations of kinematics describe the relationship between displacement, initial velocity, acceleration, and time. These relationships are crucial for determining how far the block travels up the incline after a specific initial speed, as well as its subsequent motion when forces change direction.

External Applied Forces on Inclined Planes

An external force, even if applied horizontally, influences motion along the plane when decomposed into components. This requires careful analysis to determine the effective force driving the motion along the incline, as well as its effect on the normal force and friction.

Newton's Second Law

This fundamental principle relates the net force acting on an object to its mass and acceleration. In problems like these, it is used to relate all the forces acting along the direction of the incline to the resulting acceleration of the block.

Force Decomposition on an Inclined Plane

When dealing with inclined planes, forces must be broken into components parallel and perpendicular to the plane. This includes the gravitational force, the applied horizontal force, and the normal force, each of which can be resolved into components that affect the block's motion along the incline.

Kinetic Friction

Kinetic friction is the resistive force that opposes the motion of an object sliding across a surface. The frictional force on an inclined plane is dependent on the coefficient of kinetic friction and the normal force, and it always acts opposite to the direction of movement.

Transcript

00:01 So we're going to be looking at problem 23 from chapter 5 of the physics fifth book.

00:05 The question says a 4 .8 kilogram block on a 39 degree inclined plane is acted on by a horizontal force of 46 newtons.

00:13 The coefficient of kinetic friction between the block and the plane is 0 .33.

00:18 For part a, what's the acceleration of the block if it's moving up the plane? part b, with horizontal force still acting, how far up the plane will the block go if it has an initial upward speed of 4 .3 meters per second? a propensity what happens to the block after it reaches its highest point so this is the setup we have at the moment for part a we have our object moving up some velocity v or rather acceleration but we'll deal with that later so we have some object here and it's moving up.

01:01 It's 4 .8 kilograms and we've drawn our standard force lines here.

01:08 So this is w here and this is n.

01:12 This is f.

01:13 And we have this 46 newton force, which has components 46 sine theta and 46 cos theta in these two directions, as well as our components of the weight.

01:26 And because our object is moving upwards, our friction is going downwards, so it's in the same direction as our weight.

01:36 So if we equate forces, if the block is moving up, it's ma, so this force ma, is equal to 46 cos39, which is this, this force here, minus f which is this minus mg sine 39 which is that and we know that f is equal to mute n which is equal to 0 .33 times by and now n is just going to be simply mg cos theta plus 46 sine theta so mg cos 39 plus 46 sine 39 so subbing in our values we can try and solve this and we get m .a is equal to 50 minus 15 .5 newton so a is equal to minus 3 .2 meters per second squared.

02:27 For b it asks us what is that with the horizontal force still acting how far up the plane will the block go if has an upwards initial speed of 4 .3 meters per second well we're just going to do you see about for this we want to find our s we know our u is 4 .3 meters per second we know our our final v is 0 meters per second and we know that our acceleration is negative 3 .2 meters per second.

02:55 So we have s, u, v and a.

02:57 So we use v squared as equal to u squared plus 2 as.

03:01 Now we get 4 .3 square which is u squared is equal to 2 times 3 .2 times s...

Please give Ace some feedback