asked by DungeonHunter, 8 days ago

# A block of mass 1 kg is attached to a spring with a force constant 100 N/m and rests on a roughhorizontal ground as shown in the figure. Initial displacement of block from natural length is 50 cm. Thetotal distance covered by the block if coefficient of friction between block & ground is 0.05. [g=10m/s)​

Attachments:

1

Attachments:
1

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

#SPJ2

Similar questions
Math, 4 days ago
Math, 4 days ago
Social Sciences, 8 days ago
Computer Science, 8 days ago
Geography, 5 months ago
History, 5 months ago
History, 5 months ago