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

Hey the person that did this problem got 2.4 when calculating part c for the spheres acceleration can you please show me the algebra behind it? thanks.

Explanation / Answer

a) Apply conservation of energy

final potential energy = initial kinetic energy

m*g*h = (1/2)*m*vo^2

h = vo^2/(2*g)

= 3^2/(2*9.8)

= 0.459 m

b) Again Apply conservation of energy

final potential energy = initial kinetic energy

m*g*h = (1/2)*m*vo^2 + (1/2)*I*wo^2

m*g*h = (1/2)*m*vo^2 + (1/2)*(2/5)*m*r^2*wo^2

m*g*h = (1/2)*m*vo^2 + (1/5)*m*(r*wo)^2

m*g*h = (7/10)*m*vo^2

h = 7*vo^2/(10*g)

= 7*3^2/(10*9.8)

= 0.643 m

c) acceleration of the block = -g*sin(20) - mue_k*g*cos(20)

= -9.8*sin(20) - 0.25*9.8*cos(20)

= -5.65 m/s^2

for sphere distance travelled along the ramp, d = h/sin(20)

= 0.643/sin(20)

= 1.88 m

let a is the acceleration of the sphere.

use, vf^2 - vi^2 = 2*a*d

==> a = (vf^2 - vi^2)/(2*d)

= (0^2 - 3^2)/(2*1.88)

= -2.4 m/s^2

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