### Video Transcript

A body of mass 10 root three kilograms was placed on a smooth plane inclined at 30 degrees to the horizontal. A horizontal force of 126 newtons directed toward the plane was acting on the body such that the line of action of the force and the line of greatest slope of the plane all lie in the same vertical plane. After moving for seven seconds, the body reached a speed π£, at which point the force stopped acting. And the body continued moving until it momentarily came to rest π‘ seconds after the force stopped acting. Find π£ and π‘. Take π to be equal to 9.8 meters per square second.

To answer this question, weβre going to begin by drawing a diagram. We have a smooth plane inclined at 30 degrees to the horizontal. Now, the fact that itβs smooth is important. This means there will be no frictional forces on the body. Weβre told that a body of mass 10 root three kilograms is placed on that plane. Using the formula force is equal to mass times acceleration, we find that the body exerts a downwards force on the plane of 10 root three times π, where π is acceleration due to gravity. The downward force of the body on the plane is 10 root three π newtons.

Weβre then told that a horizontal force of 126 newtons also acts on the body. And thatβs almost it for the forces acting in this diagram. We do need take into account that there is a normal reaction force of the plane on the body. This force acts perpendicular to and away from the plane as shown. So, what else do we know? Well, we know that the body was initially at rest. So, its starting speed or velocity, which we can call π’, must be equal to zero. After moving for seven seconds, it reaches a speed π£, so we can let time π‘ be equal to seven and π£ be equal to π£ meters per second.

In order to find π£, weβre going to need to establish the acceleration of the body. And so we go back to our earlier formula force is equal to mass times acceleration. We can use our diagram of forces to calculate the acceleration of the body, assuming, of course, that it moves up the plane. Now the acceleration will be parallel to the plane, essentially. So weβre going to need to work out the component of the forces that act on the body in this direction. In order to find these, weβre going to need to add two right-angled triangles to our diagrams. We add these triangles to the forces that we have. And then weβre looking to calculate the component of these forces that is parallel to the plane.

Letβs begin by looking at the horizontal force of 126 newtons. Iβve labeled the side that weβre trying to find π₯ newtons. And then we know that since corresponding angles are equal, the included angle here is 30 degrees. This is the adjacent side in our triangle. And we know that the hypotenuse is 126 newtons. Since the cosine ratio links the adjacent and the hypotenuse, we can say that cos of 30, which is adjacent over hypotenuse, will be π₯ over 126. Multiplying both sides of this equation by 126, and we find π₯ is equal to 126 times cos of 30. And so we have a force acting up the plane, letβs take that to be the positive direction, of 126 cos 30.

Weβll now go to the downward force of the weight of the body. Once again, the included angle is 30 degrees. We want to find the opposite side in this triangle, and we know the value of the hypotenuse. So, we use the sine ratio. sin π is opposite over hypotenuse. So sin 30 is π¦ over 10 root three π. If we multiply both sides of this equation by 10 root three π, we find π¦ is equal to 10 root three π times sin 30. This is acting down the plane, so we can say its force is going to be negative. This means the sum of the forces must be 126 cos 30 minus 10 root three π sin 30. This gives us a value of 14 root three or 14 root three newtons.

Now using the formula force is equal to mass times acceleration, we can calculate the acceleration during the first seven seconds. We get 14 root three equals 10 root three π, and then we divide through by root three. Finally, we divide through by 10, and we find the acceleration is 1.4 meters per square second. Now that we know this, weβre ready to calculate the value of π£ after this first seven seconds. We know acceleration is constant, so weβll use one of our SUVAT equations or equations of constant acceleration. The equation that links π£, π’, π, and π‘ is π£ equals π’ plus ππ‘. And so we get π£ equals zero plus 1.4 times seven. So, π£ is equal to 9.8 or 9.8 meters per second.

The next part of this question says that the force has stopped being applied. The body continues to move until it momentarily comes to rest π‘ seconds after the force stops acting. We need to find the value of π‘. This time we change what we know about the motion of the body. Its starting speed is 9.8, and its finishing speed π£ is zero. Weβre looking to calculate π‘, and so to do so, we need to work out the value of π, the new acceleration of the body. Since the force is no longer acting on the body, we can consider πΉ equals ππ but simply using the force of the weight of the body.

We remember that the component of the force of the body that acts parallel to the plane was 10 root three π sin 30. This is acting down the plane, so weβre going to call this force negative. We get negative 10 root three π sin 30 equals 10 root three π. And then we spot that we can divide both sides of this equation by 10 root three, meaning acceleration is negative π times sin 30. And since π is 9.8, this gives us negative 4.9 meters per square second. It makes a lot of sense that the acceleration of this body would be negative. Essentially, itβs slowing down from 9.8 meters per square second to a speed of zero. And we know that deceleration corresponds to a negative value for acceleration.

And so we go back to our earlier formula. Once again, we use π£ equals π’ plus ππ‘. But now we get an equation zero equals 9.8 minus 4.9π‘. We add 4.9π‘ to both sides. And finally, weβre going to divide by 4.9. 9.8 divided by 4.9 is two. So we find π‘ is equal to two. Under the conditions given in this question then, π£ is 9.8 meters per second and π‘ is two seconds.