Question

75. A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10,000. N/m. The spring compresses by (1) 5.5 cm (2) 2.5 cm (3) 11.0 cm (4) 8.5 cm

Answer 1

Answer:

The spring compresses by 5.5 cm. Option A is the right answer.

Explanation:

The given system has 2 kg block sliding over horizontal floor, the compression in the spring can be given by –

Given data suggests that, mass of block m = 2 kg  

 Velocity v = 4 m/ s

 Friction force = 15 N

 Spring constant = 10000 N / m

Thereby, as we know that there is an equation derived for the total work done by all the force causes change in kinetic energy, which is -

Kinetic energy change = work done by all the acting force

δK = Work done by spring + Work done by friction force

δK = Ws + Wf

$\begin{array}{l}{1 / 2 \mathrm{mv}^{2}=\mathrm{f} \text { delta } \mathrm{x}+1 / 2 \text { spring constant } \mathrm{x} \text { compression }^{2}} \\ {1 / 2 \times 2 \times 4 \times 4=15 \mathrm{x}+1 / 210000 \mathrm{x}^{2}} \\ {16=15 \mathrm{x}+5000 \mathrm{x}^{2}} \\ {5000 \mathrm{x}^{2}+15 \mathrm{x}-16=0}\end{array}$

Solving this equation, we get –

X = 5.5cm

Therefore, the compression in the spring is 5.5 cm.