A block is placed on a ramp of parabolic shape given by the equation y = x²/20 as shown in the figure. If is = 0.5, what is the maximum height above the ground at which the block can be placed without slipping?
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Hii dear,
# Answer- v = 5 m/s
# Explaination-
Since y = x^2/20 ...(1)
then dy/dx = x/10 = tanθ ...(2)
Now use a FBD to determine the equilibrium condition.
But tanθ = µ
Putting this in above eqn (2),
x/10 = µ
x = 10µ = 10×0.5
x = 5 m
Putting x=5 in eqn(1)
y = (5^2)/20
y = 1.25 m
By law of conservation of energy, gravitational energy gets converted to kinetic energy.
mgh = (1/2)mv^2
v = √(2gh)
v = √(2×1.25×10)
v = √25
v = 5 m/s
The block will move down to bottom with velocity of 5 m/s.
Hope that is useful...
# Answer- v = 5 m/s
# Explaination-
Since y = x^2/20 ...(1)
then dy/dx = x/10 = tanθ ...(2)
Now use a FBD to determine the equilibrium condition.
But tanθ = µ
Putting this in above eqn (2),
x/10 = µ
x = 10µ = 10×0.5
x = 5 m
Putting x=5 in eqn(1)
y = (5^2)/20
y = 1.25 m
By law of conservation of energy, gravitational energy gets converted to kinetic energy.
mgh = (1/2)mv^2
v = √(2gh)
v = √(2×1.25×10)
v = √25
v = 5 m/s
The block will move down to bottom with velocity of 5 m/s.
Hope that is useful...
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