A spring with spring constant of 500 N/m is being used to pull down on the hanging end of a string that is attached over a pulley to a 50 kg block that is resting on the table. If the coefficient of friction between the block and the table is .3 and the acceleration of the block is 4 m/s2 how far is the spring being stretched?

S = 0.7 m

So, the spring will be stretched as far as 0.7 m

Step-by-step explanation

Data given,

k = 500 N/m

U = 0.3

m = 50 kg

g = 10 m/sec^2

a = 4 m/sec^2

To find the spring streched 'S'

Using the net Force equation is

Fs - f = F

step 1 we need to find spring force (Fs)

Force ( spring force formula )-

Fs = k × S

step 2 we need to find friction spring (f)

f = U × N

( Where f is a fictional force, U is a coffecient of frictional force and N is normal force which is balanced by the weight of the block ).

So, N = mg ( m is a mass of block and g is a acceleration due to gravity )

f = U × m × g

step 3 we need to find net force on block (F)

F= m x a

( Where F is a net force on block, m is a mass of block and 'a' is acceleration of block )

Last step, put all equation from step 1 till step 3 into the net force equation that we have

Fs - f = F

(k × S) - ( U × m × g ) = m x a

S = ( m×a + U×m×g )/ k

= ( 50×4 + 0.3×50×10 )/500

S = 0.7 m

so, the spring will be stretched as far as 0.7 m