A block of mass m = 2.0 kg is pushed to the right with initial speed v_i = 10.0

Question

A block of mass m = 2.0 kg is pushed to the right with initial speed v_i = 10.0 m/s. Its path is on the frictionless surface except for the horizontal section BC of length L = 10.0 m where the coefficient of the kinetic friction is mu_k = 0.2. The indicated height is h = 2.0 m, the spring constant is k = 10^3 N/m. (a) What is the kinetic energy of the block at point A? (b) What is the speed of the block at point B? (c) What is the kinetic energy of the block at point C? (d) What is the maximum compression length of the spring? (e) On its return trip, where on the horizontal section to the left of point C will the block stop? What should be the initial speed of the block at which it will stop just before reaching point B?

Explanation / Answer

(a)

initial kinetic energy Ki = (1/2)*m*vi^2 = (1/2)*2*10^2 = 100 J

(b)

total energy at A = total energy at B

(1/2)*m*vi^2 = (1/2)*m*vB^2 + m*g*h

100 = (1/2)*2*vB^2 + (2*9.8*2)

vB = 7.8 m/s

------------

(c)

from B to C

Wfriction = dKE

-uk*m*g*L = (1/2)*m*(Vc^2 - vB^2)

-0.2*2*9.8*10 = (1/2)*2*(vc^2 - 7.8^2)

vc = 4.65 m/s

==================

(d)

KEC = eleastic potential energy

(1/2)*m*vc^2 = (1/2)*k*x^2

2*4.65^2 = 10^3*x^2

x = 0.21 m

=================

from c

acceleration a = -uk*g

vf = 0

vf^2 - vc^2 = 2*a*x

0 - 4.65^2 = -2*0.2*9.8*x

x = 5.51 m   from C <<<<-----answer

=================

Bonus

from work enrgy relation

EA - WBC - WCB = EB

(1/2)*m*vi^2 - uk*m*g*L - uk*m*g*L = m*g*h

(1/2)*vi^2 - 2*uk*g*L = g*h

(1/2)*vi^2 - 2*0.2*9.8*10 = 9.8*2

vi = 10.84 m/s <<<<<-------answer

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