A block of mass 2 kg is placed on the floor (μ = 0.4). a horizontal force of 7 n is applied on the block. the force of friction between the block and floor is

 The mass of the block as given in the question is 2kg.

The coefficient of static friction is 0.4.

Force applied on the block is Fnet = 2.8N

The force of friction between the block and the floor =?

Limitng friction =μN = 0.4 × 2 × 10 = 8N

As applied force is less than limiting friction

so Force of friction on block=Applied external force=7N

Getting Info...
Cookie Consent
We serve cookies on this site to analyze traffic, remember your preferences, and optimize your experience.
Oops!
It seems there is something wrong with your internet connection. Please connect to the internet and start browsing again.
AdBlock Detected!
We have detected that you are using adblocking plugin in your browser.
The revenue we earn by the advertisements is used to manage this website, we request you to whitelist our website in your adblocking plugin.
Site is Blocked
Sorry! This site is not available in your country.