If the 60 -kg skier passes point A with a speed of 5 m / s, determine his speed when he reaches

step 1 Hello and welcome to this video solution of numerate. So here we are given that there is a 150 p

If the 60 -kg skier passes point A with a speed of 5 m / s,...

Key Concepts

Force Decomposition on Inclined Planes

This involves breaking down the gravitational force into components parallel and perpendicular to an incline. The component perpendicular to the plane is balanced by the normal force, while the component parallel to the plane contributes to the acceleration of the object down the slope. This technique is a fundamental part of solving problems related to motion on an inclined plane.

Conservation of Mechanical Energy

This fundamental concept in mechanics states that in the absence of non-conservative forces (like friction), the total mechanical energy (the sum of kinetic and potential energies) of a system remains constant. It is used to relate the speed of an object at different positions along its path when there is a change in elevation, allowing one to calculate the change in kinetic energy resulting from a change in gravitational potential energy.

Energy Transformation

This concept refers to the conversion between different forms of energy as an object moves in a gravitational field. When an object descends, gravitational potential energy is converted into kinetic energy, resulting in an increase in its speed. Understanding this transformation is critical in analyzing motion on inclines or slopes.

Normal Force on Inclined Surfaces

The normal force is the force exerted by a surface perpendicular to itself in reaction to an object's weight. On an inclined surface, it is less than the full weight of the object and is computed by resolving the weight into components: the normal force equals the weight times the cosine of the inclination angle. This concept is essential when analyzing forces acting on bodies on slopes.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 So here we are given that there is a 150 pound skier at the point a and he has a speed of 5 feet per second right.

00:12 Now you determine his speed when he reaches the point b on the smooth slope.

00:17 The equation of the slope is given to be y that is equal to 50 cos of pi x by 100 right.

00:37 And you have to calculate the normal force on the skier at the b and his rate of increase in speed right.

00:49 So this is pretty simple.

00:51 So first of all what we have is this is the trajectory we will be having sort of like this right.

01:02 This is the point a at the top most section and you have got point b which is 35 feet right.

01:21 Now let's start at x equal to a which is x a that's equal to 0 feet right.

01:35 And x b is 35 feet.

01:39 This is measured along the positive direction right positive x direction.

01:45 So what you can do is you have to calculate the value of y.

01:52 So y a will be equal to if you plug in it here it's 50 cos of pi 0 over 100.

02:00 So it's pi times 0.

02:02 So it's 50 feet right.

02:04 Whereas y b if you plug in the same 50 cos of pi times 35 by 100.

02:17 Yeah that's 22 .7 feet right.

02:21 So you have got both y a and y b.

02:25 Now you have to apply the principle of conservation of energy.

02:31 So that means you have got the kinetic energy at a plus potential energy at a that's equal to kinetic energy at b right.

02:42 Assuming that assuming the height from b itself right this potential energy is calculated from b.

02:50 So a b i am taking right.

02:54 Ok this is half m v square plus m g.

03:04 Let's say here i take y a minus y b half m v square right.

03:15 Mass is you have got 150 over 32 .2 the value of g itself right.

03:25 You is 5 square that's given right plus this is again 150 over here only you have got 150 right...

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