A block of Mass M travels over a frictionless horizontal surface at 3.00m/s. The

Question

A block of Mass M travels over a frictionless horizontal surface at 3.00m/s. The block encounters frictionless incline as shown in Figure below.

a) What vertical distance up the incline does the block travel before coming momentarily to rest?

b) Suppose a solid sphere of identical mass is rolling without slipping a rough surface and encounters a rough incline. If the velocity of the center of mass of the sphere is the same as that of the block how far up the incline does it roll?

c) Determine the acceleration of the blcok and sphere objects if the angle is 20 degrees and kinetics' friction coefficent µ=.25

please show me all steps

Explanation / Answer

a)

v = speed at bottom of incline = 3 m/s

h = vertical height

using conservation of energy

Potential energy at the top of incline at height "h" = kinetic energy at the bottom

mgh = (0.5) m v2

h = v2/2g

h = (3)2 /(2 x 9.8) = 0.46 m

b)

v = speed at bottom of incline = 3 m/s

w = angular speed

h = vertical height

using conservation of energy

Potential energy at the top of incline at height "h" = kinetic energy at the bottom + rotational KE at bottom

mgh = (0.5) m v2 + (0.5) Iw2

mgh = (0.5) m v2 + (0.5) (0.4)(m r2) (v/r)2 = (0.7)m v2

h = (0.7)v2/2g

h = (0.7)(3)2 /(2 x 9.8) = 0.32 m

c)

for the block :

f = frictional force

net force = mg Sin20 + f = mg Sin20 + µmg Cos20

ma = mg Sin20 + µmg Cos20

a = g Sin20 + µg Cos20

a = (9.8) Sin20 + (0.25) (9.8) Cos20 = 5.65 m/s2

for the sphere :

Torque equation is given as

f r = Ia/r

f = (0.4) (mr2) (a/r2)

f = (0.4) ma

force equation is given as

net force = mg Sin20 + f

ma = mg Sin20 + (0.4) ma

a = 2.4

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