A dog with mass M has its string attached to one end of a spring which runs without friction along horizontal overhead rod. The other end of the spring is fixed to a wall. The spring constant is k. The string is massless and inextensible and it maintains a constant angle θ with the overhead rod, even when the dog moves. There is friction with coefficient μ between the dog and the ground. 5x cm is the maximum distance that the dog moving slowly can stretch the spring beyond its natural length. Find x. Use: M = 30 kg, θ = 37°, k = 400 N/m and μ = 1/3
Physics · Mixed Chapter
JEE Mains

Question

A dog with mass M has its string attached to one end of a spring which runs without friction along horizontal overhead rod. The other end of the spring is fixed to a wall. The spring constant is k. The string is massless and inextensible and it maintains a constant angle θ with the overhead rod, even when the dog moves. There is friction with coefficient μ between the dog and the ground. 5x cm is the maximum distance that the dog moving slowly can stretch the spring beyond its natural length. Find x. Use: M = 30 kg, θ = 37°, k = 400 N/m and μ = 1/3

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Let string tension be T. It makes θ=37° below the horizontal from the spring end to the dog. The horizontal spring force is kΔl=T cosθ. Step 2: At the limiting position for slow motion, horizontal friction is f=μN=T cosθ, while vertical equilibrium gives N+T sinθ=Mg, so N=Mg−T sinθ. Step 3: Therefore T cosθ=μ(Mg−T sinθ), or T(cosθ+μsinθ)=μMg. With μ=1/3, M=30 kg, g=10, sin37°=3/5 and cos37°=4/5, we obtain T=100 N. Step 4: kΔl=T cosθ=100×4/5=80 N, so Δl=80/400=0.20 m=20 cm. As Δl=5x cm,…