Answer:
d= 4.38 m
Explanation:
Principle of conservation of energy :
E1=E2 Formula (1)
E1: Total energy at the top of the ramp
E2: Total energy in the final part of the ramp.
Total energy of the block is conserved because there is no friction from 1 to 2.
E = K + U
K : Kinetic energy in Joules (J)
U: Gravitational potential energy in Joules (J)
K = (1/2) *m*v²
U= m *g*h
Where:
m: block mass (kg)
v: speed of the blck
g= the acceleration of gravity (m/s²)
h= is the heigt measured from the floor (m)
Calculating of the speed of the block at the end of the ramp
Data
v₁ =0
h₁ = 3.42 m*sin (52.3)°= 2.706 m
h₂ = 0
We replace data i the formula (1)
E1=E2
K₁ + U ₁= K ₂ + U ₂
(1/2) *m*v₁² + m*g*h₁ = (1/2) *m*v₂² + m*g*h₂
0+ m*(g)*(h₁) = (1/2) *m*v₂² + 0 we divide by m on both sides of the equation
(g)*(h₁) = (1/2)*v₂²
[tex]v_{2} = \sqrt{2gh_{1} }[/tex]
[tex]v_{2} = \sqrt{2*(9.8)* (2.706 ) }[/tex]
v₂ = 7.28 m/s
Newton's second law for the block that slides horizontally along the concrete:
∑F = m*a
f: fricition force , f= μk*N N: normal force
N = W = m*g
f= (0.618) * m*g
-f = m*a
-(0.618) * m*9.8 = m*a we divide by m on both sides of the equation
a = - 6.056 m/s²
Kinematics for the block that slides horizontally along the concrete:
Because the block moves with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = v₂ = 7.28 m/s
vf = 0
a = - 6.056 m/s²
We replace data in the formula (2)
vf²=v₀²+2*a*d
0= (7.28)²+2* (-6.056 )*d
2*(6.056 )*d =(7.28)²
d =(7.28)² / (12.112)
d= 4.38 m
