A small object sits on a compressed spring. The spring will uncompress and prope

Question

A small object sits on a compressed spring. The spring will uncompress and propel the obejct forward. There is a short patch over which the object feels friction with the ground. The rest of its travel is frictionless. The object finally travels up a long ramp

-The force of gravity is 9.81m/s2 and it is pointed toward the center of the earth

The object has a mass of 2kg

The spring constant is 300 N/m

The spring is intially compressed by .3m

The small region in which the object experiences friction is 1m wide and has a coefficient of kinetic friction of 0.06 the rest of the setup is frictionless.

A. What is the veloctiy of the object just after it leaves the spring and before it reaches the rough region? Solve using conservation of energy without any shotcuts.

B. What is its velocity after it has passed through the rough region? Solve using conservation of energy without any shortcuts.

C. What is the maximum height it reaches up the ramp? solve using conservation of Energy, without any shortcuts.

Explanation / Answer

A) gain in kinetic energy = loss of elastic potential energy

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

v^2 = k*x^2/m

v = sqrt(k*x^2/m)

= sqrt(300*0.3^2/2)

= 3.67 m/s <<<<<<<-----------------Answer

B) workdone friction = change in kinetic energy

Fk*d*cos(180) = (1/2)*m*(vf^2 - vi^2)

-mue_k*m*g*d = (1/2)*m*(vf^2 - vi^2)

-2*mue_k*g*d = vf^2 - vi^2

==> vf = sqrt(vi^2 - 2*mue_k*g*d)

= sqrt(3.67^2 - 2*0.06*9.81*1)

= 3.51 m/s

C) final gravitational potential energy = initial kinetic energy

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

h = vf^2/(2*g)

= 3.51^2/(2*9.81)

= 0.628 m

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