Question

A 4 kgi banana bunch is suspended motionless from a spring balance which force constant is 300 Newton per metre the length of the which is Spring will be stretched

Answer 1

Answer:

According to the question,

The maximum mass scale can read M=50kg

The maximum displcement of spring =

length of scale, l=20cm=0.2m

T (time period) =0.6s

Maximum force exerted on the spring F=mg

So,

g=acceleration due to gravity =9.8m/s

2

F=50×9.8=490

Since, speing constant,

k=

l

F

=

0.2

490

=2450Nm

−1

m is suspended from the balance

T=2π

k

m

Since,

m=(

2π

T

)

2

×k

=(

2×3.14

0.6

)

2

×2450

≈22kg