Question

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.

Answer 1

The potential energy of the block-Earth system as a function of x is U(x) = mgxsin(θ). The speed of the block at the bottom of the incline is v_x,f = sqrt(2gxsin(θ)).

Step-by-step explanation:

To derive an expression for the potential energy of the block-Earth system as a function of x, we first need to consider the gravitational potential energy, which is given by U(x) = mg. In this case, the height h is equal to sin (θ). Therefore, the potential energy can be expressed as U(x) = mixin (θ).

To determine the speed of the block at the bottom of the incline, we can use the conservation of mechanical energy. At the top of the incline, the potential energy is zero, and at the bottom, all the potential energy is converted into kinetic energy. Therefore, we can equate the potential energy and kinetic energy, resulting in mixing (θ) = 1/2mvx,f2. Rearranging the equation, we find that the final speed of the block at the bottom of the incline is given by vx,f = sqrt(2gxsin(θ)).

Answer 2

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Step-by-step explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

UG = UK

- Where UK is kinetic energy given by:

UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by:

Vx,f = sqrt (2*g*x*sin(Q))