Question

A block of mass 1 kg is dropped on a
spring from a height of 20 cm. If the spring
compresses by 5 cm, the force constant
of the spring is (g = 10 ms)
1) 2000 Nm
2) 1000 Nm
3) 200 Nm
4) 500 Nm
​

Answer 1

Answer:

Use energy conservation

$mgh = \frac{1}{2} k {x}^{2} \\ 1 \times 10 \times (20 + 5) \times \frac{1}{100} = \frac{1}{2} \times k \times {5}^{2} \times \frac{1}{ {100}^{2} } \\ k = 2000$

Answer 2

Answer:

K=2000NM^-1

Explanation:

Gravitational capability electricity is the electricity possessed or received with the aid of using an item because of a extrade in its function whilst it's far found in a gravitational field. In easy terms, it could be stated that gravitational capability electricity is an electricity this is associated with gravitational pressure or to gravity.

Gravitational capability electricity of the falling mass transformed into capability electricity of spring (as after compression spring is at rest). Therefore we will write

U =1/2 kx²

mg(h+x)=1/2 kx²

1 kg. 10 .m/s²(20+5/100 m)=1/2 k (5/100 m )²

2.5kg m²/s²=1/2 k(0.05 m )²

K=200 kg / s²

K=2000NM^-1

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