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4 years ago

14

Eight joules of work is required to stretch a spring 0.5 meter from its natural length. Find the work required to stretch the sp

ring an additional 0.40 meter.

1 answer:

Alexxandr [17]4 years ago

8 0

Additional Work done to stretch the spring 0.4m is 6.4J

<u>Explanation:</u>

<u></u>

Given -

Work done to stretch a spring, W1 = 8J

Distance , d = 0.5m

Distance2, x = 0.4m

Work done 2, W2 = ?

Using Hooke's law,

Work done = k x

where k is spring constant

For distance 0.5m :

8 J = k X 0.5m

k = 8/0.5 = 16 J/m

For distance 0.4m:

W2 = 16 X 0.4 J

W2 = 6.4J

Additional Work done to stretch the spring 0.4m is 6.4J

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3 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

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At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

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