Question

A mass of 1 kg is dropped from a height of 2 m on a horizontal spring board. The vertical spring supporting the board has a spring constant of 87.5 N/m. The maximum distance by which the mass compresses the spring is close to

Answer 1

mass of object = 1 kg 
initially the block is at rest the spring is relaxed so we se the only energy is present is only the gravitational potential energy which means U1=mg*hight 
when overall the block falls then it compress the spring which means 
hight will become (h+x) 
so we can say that U1= mg(h+x) 
when the block will be dropped only the final potential energy will be the spring potential energy dude :) 
which means U2= 1/2 kx^2 
by conservation of energy we can say that dude 
U1=U2
1/2kx^2= mg(x+h) putting values as m=1kg k=87.5N/m^2  and g = 10 m/s^2 and h = 2m  but first i will solve this equation before putting values to make this equation shorter and fit for easy calculation 
1/2 kx^= mg(x+h)
kx²=2mg(x+h)
kx²-2mgx-2mgh = 0
now dude it is quadratic equation in x 
dude now using quadratic formula we can find its root and the value of x will be maximum 
as a=k b= -2mg and c= -2mgh 
now using quadratic formula 
x= (-b + ,- √b²-4ac)/2a
x=(+2mg+√(2mg)² +4k(2mgh )/2k
now i will directly put the values of k m h and g 
x= [(2*1*10 + √4*10^2*1 + 4*87.5(2*10*2*1)] /87.5*2
x=[20 +√400+1400]/175
x=62.42/175 = 0.4 approx 
my answer is correct mark it as the brainliest one i just spent 15 minutes in it because of calculations 

Answer 2

a)0.8m

We will apply the Principle of Conservation of Mechanical Energy.

We will take the final  position of mass m as the reference configuration for gravitational potential energy.

Hence:-

 Ui+Ki=Uf+Kf

 Since, Ki=Kf=0

Hence,

Ui=Uf

⟹mg(h+x)=21kx2

⟹10(2+x)=287.5x2

⟹87.5x2−20x−40=0

x=2×87.520+202+4×87.5×40 neglecting −ve value of x.

⟹x=0.8m

Hence, answer is option-(A).

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