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A block of mass 1 kg is attached to a spring with a force constant 100 N/m and rests on a rough

horizontal ground as shown in the figure. Initial displacement of block from natural length is 50 cm. The

total distance covered by the block if coefficient of friction between block & ground is 0.05. [g=10m/s)

## Answers

### Answer : x=25m

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**25 m is the total distance covered by the block if the coefficient of friction between the block & ground is 0.05**

**Given**,

Mass of the block= 1 kg

Force constant= 100N/m

Initial displacement of block from natural length = 50 cm

**To find**,

The total distance covered by the block if the coefficient of friction between the block & ground is 0.05

**Solution**,

According to Hooke's law, the applied force F equals a constant time the displacement or change in length, "x".

i.e., F=kx

Here,

- F is the restoring force exerted by the spring

- k is the constant factor characteristic of the spring. The value of k depends on the kind of elastic material under consideration, its dimensions, and its shape.

- x is the displacement from its position at the equilibrium

F = kx

F = 100 * 0.5

F = 50 N

Now,

f = μmg

f = 0.05 * 1 * 10

f = 0.5 N

Work done by friction = μmg*x

Initial Potential energy =

Equating both equations,

μmg*x =

0.5 * x = 1/2 * 100 * 1/2 * 1/2

x = 25 m

**25 m is the total distance covered by the block if the coefficient of friction between the block & ground is 0.05**

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