Question

A vertical spring of force constant 100 N/m is attached with a hanging mass of 10 kg. Now an external force is applied on the mass so that the spring is stretched by additional 2 m. The work done by the force F is : (g = 10 m/s²)

• A. 200 J
• B. 400 J
• C. 450 J
• D. 600 J

Answer 1

Answer: (A)

200 J

Answer 2

At equilibrium , mg = kx₀ Ž x₀ = $\frac { \mathrm { mg } } { \mathrm { k } }$ = 1m

\ W_ext = U₂ – U₁

=$\frac { 1 } { 2 } \mathrm { kx } _ { 1 } ^ { 2 }$ + mgh]

= $\frac { 1 } { 2 }$ k ( – $\mathrm { x } _ { 1 } ^ { 2 }$ ) – mgh

=$\frac { 1 } { 2 }$ × 100 × (3² – 1²) – 10 × 10 × 2 = 200 J