AP EAPCET 18 May 2024 Shift 1 Solved Paper

Spring constant, k=5×103N∕m
Initially stretched position, x1=10cm=0.1m
Further stretched position, x2=20cm=0.2m
Work Done:
The work done to further stretch the spring from x1 to x2 is equal to the change in potential energy of the spring. This can be calculated using the formula:
‌ Work done ‌=‌

1

2

k[(x2)2−(x1)2]
Substitute the given values:
‌=‌

1

2

×5×103[(0.2)2−(0.1)2]
‌=‌

1

2

×5×103×(0.04−0.01)
‌=‌

1

2

×5×103×0.03
‌=2.5×103×0.03
‌=75N⋅m
Therefore, the work required to stretch the spring further by another 10 cm is 75N⋅m.