Question

A 10kg block is Pulled along a horizontal
Surfact by a fores of son at an ansie
of 37° with the horizontal If the
coefficient of sliding friction 6/n the
block ound the surface is 0.2
(8-10 mis? Sin 37=016 and cos 37.2008)
of what frictional for acting on the block
b what is the acceleration of the block?​

Answer 1

Answer:

Explanation:

I hate these kinds of problems,  luckily I can't understand how much the kinetic friction is for this ,  the words are all mixed around.  and don't read well.  Maybe this went through a translator program?   My suggestion draw the free body diagram.   so you can see where the forces are, and how they are acting.   getting the free body diagram right.. usually makes these problems pretty straight forward.   just do the steps and you get the answer.

Answer 2

Answer:

I cant understand the question well but i will tell u how to solve it

first apply f = ma eqaulation vertically...then u will get an r = mg which means normal upward force = weight of the object (to find weight multiply mass by 9.81)

$f = ma \\ r - mg = 0 \\ r = mg$

after that use friction eqaution which is

$f = \mu \times r$

(mu) is the coefficient of friction and (r) is the vertical normal force then by substituting value you can get an answer for the friction

the use

$f = ma \\ f \cos(37) - fr = ma$

by using the above eqaution you can get an answer for acceleration...

hope you can understand...