Answer:
0.57 m
Explanation:
Consider the motion while crossing the rough patch
v₀ = initial velocity of the block at the start of rough patch
v = final velocity of the block after crossing the rough patch = 3.2 m/s
μ = coefficient of kinetic friction = 0.32
acceleration of the block due to frictional force is given as
a = - μg
a = - (0.32) (9.8)
a = - 3.136 m/s²
d = width of the rough patch = 15 cm = 0.15 m
Using the equation
v² = v₀² + 2 a d
3.2² = v₀² + 2 (- 3.136) (0.15)
v₀ = 3.34 m/s
m = mass of the block = 1.5 kg
h = height of ramp
Using conservation of energy
Gravitational potential energy at the top = Kinetic energy at the bottom of the ramp
m g h = (0.5) m v₀²
g h = (0.5) v₀²
(9.8) h = (0.5) (3.34)²
h = 0.57 m