Question

A 0.5kg block slides from the point A on a horizontal track with an initial speed 3 m//s towards a weightless horizontal spring of length 1m and force constant 2N//m. The part AB of the track is frictionless and the part BC has the confficient of static and kinetic friction as 0.20 respectively. If the distancences AB and BD are 2m and 2.14m respectively, find total distance through which the block moves before it comes to rest completely. (g=10 m//s^(2) ).

Answer 1

Given:

1. Mass of block, m = 0.5 kg
2. Initial speed, u = 3m/s
3. Length of spring = 1m
4. Force constant, k = 2N/m
5. Coefficient of friction on part BC of track, μ = 0.2
6. Distance AB = 2m
7. Distance BD = 2.14m

To find:

1. Total distance through which the block moves before it comes to rest

Answer:

• Track AB is frictionless, this implies that the block moves along this distance without any loss in the initial kinetic energy.
• Initial kinetic energy of block = 1/2 m*u² = 1/2 0.5*3² = 2.25J
• In the path BD as friction is present
• Work done against friction = μmgs =  μmg(BD) =0.2×0.5×10×2.14=2.14J

• At D the KE of the block is=2.25-2.14=0.11J
• Let us assume that the spring is compressed by x
• 0.11= 1/2 ×k×x² +μ kmgx
• 0.11= 1/2 ×2×x² +0.2×0.5×10x
• x² + x−0.11=0
• x=0.1 m
• Thus, after moving a distance of 0.1 m the block comes to rest.
• So, the total distance moved by the block is -
• =AB+BD+ x
• =2+2.14+0.1
• =4.24m

Total distance through which the block moves before it comes to rest =4.24m.