Consider the image below. A 0.50kg block is shown at various different points in
ID: 1371344 • Letter: C
Question
Consider the image below. A 0.50kg block is shown at various different points in time. Note that there is onle one block. The block starts compressed 0.15m on a spring with spring constant: k = 3900 N/m and travels on a surface, down a small hill, until it comes to rest at point D. The surface is frictionless from point A to point B, from point B to point D there is a coefficient of friction = 0.37. If the height of the hill h = 0.75 m, find the following:
a) what is the velocity v2?
b) What is the distance from point B and point D (labeled S)?
c) What is the velocity of the block at point C, 1/2 way between B and D?
Given: v1 = 13. 25 m/s
Explanation / Answer
(a)
Initial Kinetic Energy of Block = 0.5* m*v1^2
Initial Potential Energy of Block = m*g*h
Now, At Point B
Initial Kinetic Energy of Block = 0.5* m*v2^2
Using Energy Conservation
Initial Energy = Final Energy
0.5* m*v1^2 + m*g*h = 0.5* m*v2^2
0.5*13.25^2 + 9.8*0.75 = 0.5*v2^2
Solving for v2
v2 = 13.8 m/s
Velocity v2 = 13.8 m/s
(b)
Coefficient of friction = 0.37
Initial velocity = 13.8 m/s
Final Velcoity = 0
Distance = S
Decceleration, a = F/m
Where , F = u*N
F = 0.37 * m*g
a = 0.37*g
Now,
v^2 = u^2 - 2*a*S
0 = 13.8^2 - 2*0.37*9.8 * S
S = 26.26 m
Distance from point B and point D, S = 26.26 m
(c)
Distance d = 26.26/2 = 13.13 m
Deccelartion , a = 0.37*g = 0.37*9.8 = 3.626 m/s^2
Initial Velocity u = 13.8 m/s
Final Velocity v = ?
v^2 = u^2 - 2*a*d
v = sqrt(13.8^2 - 2*3.626*13.13) m/s
v = 9.8 m/s
Velocity of the block at point C , v = 9.8 m/s