A 40 kg boy jumps from a height of 4 m onto a plate-form mounted on springs.As the spring compress, the plate-form is pushed down a maximum distance of 0.2 cm below it's initial position, and it then rebounds .What is the loss in potential energy by the boy?
Answers
Answer:
The plate and spring play no part in the problem. The potential energy "loss" is just M G H where M is the mass, G is gravitation acceleration (9.8m/sec^2) and H the height (4 meters).
Explanation: Whether the spring system "loses" energy, cannot be determined from the information given.
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^2mgx−
2
1
kx
2
=0−
2
1
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)40(10)(0.20)−
2
1
k(0.2)
2
=0−
2
1
(40)(2×10×4)
so we have
80 - 0.02 k = -160080−0.02k=−1600
k = 84000k=84000
now again by energy equation
mgx - \frac{1}{2}kx^2 = \frac{1}{2}mv_f^2 - \frac{1}{2}mv^2mgx−
2
1
kx
2
=
2
1
mv
f
2
−
2
1
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(10)(0.1)−
2
1
(84000)(0.10)
2
=
2
1
(40)v
f
2
−
2
1
(40)(2(10)(4))
40 - 420 = 20v_f^2 - 160040−420=20v
f
2
−1600
v_f = 7.8 m/sv
f
=7.8m/s