Answer:
E) 80 N/m
Explanation:
Given;
mass of the block, m = 4.8 kg
displacement of the block, x = -0.5 m
velocity of the block, v = -0.8 m/s
acceleration of the block, a = 8.3 m/s²
From Newton's second law of motion;
F = ma
Also, from Hook's law;
F = -Kx
where;
k is the force constant
Thus, ma = -kx
k = -ma/x
k = -(4.8 x 8.3) / (-0.5)
k = 79.7 N/m
k ≅ 80 N/m
Therefore, the force constant of the spring is closest to 80 N/m
here we can say that net force on the student vertically upwards will be counter balance by his weight downwards
Let net force F is exerted by each hand
so here we will have
so the force exerted by each hand will be 414.1 N
I think the answer you're looking for is answer B) Not there. Good luck and hoped it helped!
