P12: A 2.0-kg block is held at rest against a spring of constant 2700 N/m, compr
ID: 1778384 • Letter: P
Question
P12: A 2.0-kg block is held at rest against a spring of constant 2700 N/m, compressing it 5.5 cm. The block and the spring are located on a horizontal surface. When the system is released, the block slides down a ramp with an angle of 35° to a second horizontal surface 1.7 m below the first. The incline has a coefficient of kinetic friction of 0.29; the horizontal surfaces are frictionless. Please find a) the potential energy stored in the spring before it is released. b) the mechanical energy lost by the block once it reaches the lower horizontal surface. c) the final speed of the block as it slides along the lower horizontal surface.Explanation / Answer
a)
potential energy stored Ue = (1/2)*k*x^2 = (1/2)*2700*0.055^2 = 4.1 J
(b)
on the incline
frictional force fk = uk*m*g*costheta
work done by frictional force Wf = -fk*L = -fk*H/sintheta
Wf = -uk*m*g*costheta*H/sintheta = -uk*m*g*H/tantheta
energy lost by block E = Wf = -0.29*2*9.8*1.7/tan35 = -13.8 J
(c)
initial energy at the top Ei = m*g*H + (1/2)*k*x^2 = (2*9.8*1.7) + (1/2)*2700*0.055^2
wf = -13.8 J
final energy Ef = (1/2)*m*v^2
Wf = Ef - Ei
-13.8 = (1/2)*2*v^2 - (37.4)
v = 4.86 m/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.