Question

. A 40kg boy jumps from a height of 4m on to a plate form
mounted on springs. As the springs compress, the plate form is pushed down a maximum distance of 0.2m below its initial position, and it then rebounds. What is the boy's speed at the instant the plate form is depressed 0.1m. [Ans: 7.9m/s 1​

Answer 1

Answer:

Final speed of the platform at x = 0.1 m is 7.8 m/s

Explanation:

As we know that the platform will perform SHM

here we know by energy conservation

$mgx - \frac{1}{2}kx^2 = 0 - \frac{1}{2}mv^2$

so we have

$40(10)(0.20) - \frac{1}{2}k(0.2)^2 = 0 - \frac{1}{2}(40)(2\times 10\times 4)$

so we have

$80 - 0.02 k = -1600$

$k = 84000$

now again by energy equation

$mgx - \frac{1}{2}kx^2 = \frac{1}{2}mv_f^2 - \frac{1}{2}mv^2$

$40(10)(0.1) - \frac{1}{2}(84000)(0.10)^2 = \frac{1}{2}(40)v_f^2 - \frac{1}{2}(40)(2(10)(4))$

$40 - 420 = 20v_f^2 - 1600$

$v_f = 7.8 m/s$

#Learn

Topic : Energy conservation

https://brainly.in/question/11826049

Answer 2

Solution is on the photos