Flag Mechanics> A spring of spring constant 5 × 10 3 N/m ...
question mark

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 is
a) 12.50 N-m
b) 18.75 N-m
c) 25.00 N-m
d) 6.25 N-m

satendra , 11 Years ago
Grade 12th pass
anser 4 Answers
Vasantha Kumari

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

Last Activity: 11 Years ago
Vaishnavi
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
             
Last Activity: 7 Years ago
Sohail
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
 
Last Activity: 7 Years ago
Kushagra Madhukar
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
Last Activity: 5 Years ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments