A 2.5-kg block initially at rest is released from a spring of force constant 900

Question

A 2.5-kg block initially at rest is released from a spring of force constant 900 NIm compessed by 50 cm and encounters two hills, as shown below. The height of each hill is h = 4.0 m, and the length of the horizontal ground between them is also h4.0 m. All surfaces are frictionless except the ground between the hills. a What is the coefficient of kinetic friction of the ground between the hills if the block comes to rest at the top of the right hill? that the top of each hill forms a circular are of radius h. Find the maximum ssion of the spring if the block is not to lose contact with the top of the left hill. (b) Imagine

Explanation / Answer

(A) Applying work - energy theorem,

total work done = change in KE

Work done by spring force + work done by gravity + work done by friction = Kf - Ki

k x^2 /2 - m g h - f h = 0 - 0

(900)(0.50^2)/2 - (2.5 x 9.8 x 4) = f (4)

f = 3.625 N

and f = uk N = uk m g

uk = (3.625)/(2.5 x 9.8)

uk = 0.15

(B) at left hill,

m g - N = m a_c = m v^2 / r

N = m g - m v^2 / h = 0

v = sqrt(g h )

Applying energy conservatin,

PEi + KEi = PEf + KEf

(900 x^2 / 2 + 0 ) + (0) = (m g h + 0 ) + m v^2 /2

450 x^2 = (2.5 * 9.8 * 4) + (2.5 * 9.8 * 4/ 2)

x = 0.572 m Or 57.2 cm  

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