Answer:
UG (x) = m*g*x*sin(Q)
Vx,f (x)= sqrt (2*g*x*sin(Q))
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))
