A1.5-kg block at rest on a ramp of height h. when the block is released, it reaches the bottom of the ramp and moves across a surface that is frictionless except for one section of width 15 cm that has a coefficient of kinetic friction µk=0.32. find h such that the block's speed after crossing the rough patch is 3.2 m/s.

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

the assestment will probley apele with cause and effect

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j