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A spring of spring constant 5 × 103 N/m is stretched initially by 5cm from the unstretched position. Then the work required to stretch it further by another 5 cm isa) 12.50 N-mb) 18.75 N-mc) 25.00 N-md) 6.25 N-m

satendra , 10 Years ago
Grade 12th pass
anser 4 Answers
Vasantha Kumari

Last Activity: 10 Years ago

To calculate the work required to stretch the spring further by another 5 cm, we will use the formula for the work done in stretching a spring:

W = (1/2) k (x2² - x1²)

Where:

W = Work done
k = Spring constant (given as 5 × 10³ N/m)
x1 = Initial displacement (5 cm = 0.05 m)
x2 = Final displacement (Initial displacement + additional displacement = 0.05 m + 0.05 m = 0.10 m)
Step 1: Write the formula
W = (1/2) × k × (x2² - x1²)

Step 2: Substitute the values
W = (1/2) × 5000 × (0.10² - 0.05²)
W = 0.5 × 5000 × (0.01 - 0.0025)
W = 2500 × (0.0075)

Step 3: Calculate
W = 2500 × 0.0075
W = 18.75 N-m

Final Answer: (b) 18.75 N-m

Vaishnavi

Last Activity: 6 Years ago

Here, W=1/2k(x22-x12)
               = 1/2×5×103((10/100)2-(5/100)2)
               =1/2×5×103/104(100-25)
               =18.75J
Hence the work required to stretch the spring further by another 5cm is 18.75 N-m
Thanks and regards
             

Sohail

Last Activity: 6 Years ago

Work done by spring=1/2k(x22-x12)
K=5×103N/m
X2=10×10-2m
X1=5+10-2m
W=1/2×5×103(100-25)10-4
W=18.75N-m
 

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the attached solution to your problem.
 
W = 0.5k(x22 – x12)
               = 0.5 × 5 × 103((10/100)2 - (5/100)2)
               = 0.5 × 5 × 103/104 (100-25)
               =18.75 J
Hence the work required to stretch the spring further by another 5cm is b) 18.75 N-m
 
Hope it helps.
Thanks and regards,
Kushagra

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