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Vikki [24]
3 years ago
5

. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of

friction for the lawn is 0.55. How much force is required to start sliding the wrapped leaves
Physics
1 answer:
Rina8888 [55]3 years ago
7 0

The force required is 319 N

Explanation:

The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:

F_f = \mu W

where

\mu is the coefficient of friction

W is the weight of the object

Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

\mu = 0.55 (coefficient of friction)

W=580 N (weight of the leaves)

Substituting, we find:

F=(0.55)(580)=319 N

Learn more about force of friction:

brainly.com/question/6217246

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#LearnwithBrainly

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