Question

The horizontal surface on which the block slides is frictionless.The speed of the block before it touches the spring is 6.0 m/s. Howfast is the block moving at the instant the spring has beencompressed 15 cm? k = 2.0 kN/m

Question 7 answers

3.7m/s

4.4m/s

4.9m/s

5.4m/s

14m/s

Answer 1

Answer:

V = 5.4 m/s

Explanation:

It is given that,

Let mass of the block, m = 10 kg

Spring constant of the spring, k = 2 kN/m = 2000 N/m

Speed of the block, v = 6 m/s

Compression in the spring, x = 15 cm = 0.15 m

Let V is the speed of the block moving at the instant the spring has been compressed 15 cm. It can be calculated using the conservation of energy of spring mass system.

$\dfrac{1}{2}m^2=\dfrac{1}{2}kx^2+\dfrac{1}{2}mV^2$

$mv^2=kx^2+mV^2$

$V^2=\dfrac{mv^2-kx^2}{m}$

$V^2=\dfrac{10\times 6^2-2000\times (0.15)^2}{10}$

V = 5.61 m/s

From the given options,

V = 5.4 m/s

Hence, this is the required solution.