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2 years ago

8

You attach a 1.10 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by

0.200 m and release it from rest. Assume the block slides on a horizontal surface with negligible friction. The block reaches a speed of zero again 0.100 s after release (for the first time after release). What is the maximum speed of the block (in m/s)?

1 answer:

dangina [55]2 years ago

6 0

Answer:

6.3 m/s

Explanation:

m = mass of the block = 1.10 kg

k = spring constant of the spring

x = stretch in the spring = 0.2 m

t = time taken by block to come to zero speed first time = 0.100 s

T = Time period of oscillation

Time period of oscillation is given as

T = 2t

T = 2 (0.1)

T = 0.2 s

Time period is also given as

$T = 2\pi \sqrt{\frac{m}{k}}$

$0.2 = 2(3.14) \sqrt{\frac{1.10}{k}}$

k = 1084.6 N/m

v = maximum speed of the block

using conservation of energy

Maximum kinetic energy = Maximum spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(1.10) v² = (1084.6) (0.2)²

v = 6.3 m/s

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