Question

1 Warm-Up Questions 1 A block of mass m = 3 kg is dropped from the top of a half-pipe (half-circle) of radius R = 2 m starting from rest. The surface of the incline does have friction. If the block comes to rest at half of its initial height, how much work was done by friction? Work: 2 Initially, a ball of mass 0.4 kg is compressed by a spring, with spring constant k = 980 N/m, by a distance d from its rest length and released until the ball reaches a maximum height of 5+d meters. - a) Assuming energy is conserved, what is the initial compression distance, d, of the spring? Hint: use the positive value of the quadratic formula. -- WNNNM b) If energy is not conserved (due to friction, air resistance, etc), the ball reaches a maximum height of 4+ d meters. How much energy is lost due to non-conservative work?

Answer 1

(1)

The initial and final condition of the block is at rest so the work done by the friction force should be equal to the change in potential energy of the block.

The initial height of the block is equal to the radius of the semicircle.

Work done by friction force = change in potential energy of the block

= mghi - mghf

= 3*9.81(2-1)

Work done by friction force = 29.43 N

(2)

(a)

Apply the conservation of energy.

The potential energy of the ball = potential energy of the spring

mgh = 0.5kd2

0.4*9.81*5 = 0.5*980*d2

d = 0.2 m

(b)

The energy lost due to non-conservative work will be equal to the difference between the potential energy of the block in case of a conservative force and a non-conservative force system.

the energy lost due to non-conservative work = (PEb)conservative force - (PEb)Non-conservative force

= mghc - mghnc

=0.4*9.81(5+d -4-d)

the energy lost due to non-conservative work = 3.924 J

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