A horizontal spring attached to a wall has a force constant of k = 820 N/m. A bl

Question

A horizontal spring attached to a wall has a force constant of k = 820 N/m. A block of mass m = 1.30 kg is attached to the spring and rests on a horizontal, frictionless surface as in the figure below.

(a) The block is pulled to a position xi = 5.00 cm from equilibrium and released. Find the elastic potential energy stored in the spring when the block is 5.00 cm from equilibrium and when the block passes through equilibrium.

(b) Find the speed of the block as it passes through the equilibrium point.
m/s

(c) What is the speed of the block when it is at a position xi/2 = 2.5 cm?
m/s

(d) Why isn't the answer to part (c) half the answer to part (b)?

Explanation / Answer

k = 820 N/m ; m = 1.3 kg

a)x = 5 cm = 0.05 m

PE of spring is given by

PE = 1/2 k x^2

PE = 1/2 x 820 x 0.05^2 = 1.025 J

Hence, PE = 1.025 J

for x = 0 , PE = 1/2 k x^2 = 0

b)from conservation of energy

1/2 m v^2 = 1/2 k x^2 = 1.025

v = sqrt (2 x 1.025/1.3) = 1.26 m/s

Hence, v = 1.26 m/s

c)when x = 2.5 cm = 0.025, PE is

PE = 1/2 x 820 x 0.025^2 = 0.256 J

v = sqrt (2 x 0.256/1.3) = 0.63 m/s

Hence, v = 0.63 m/s

d)Because,

1/2 m v^2 = 1/2 k x^2

v = x sqrt (k/m)

speed has direct relation to the displacement, if its halved, speed will also be halved.

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