Question

A box of mass 20 kg is pushed along a rough floor with a velocity 2 m/s and then let go. The box moves 5m

on the floor before coming to rest. What must be the frictional force acting on the box?

can someone give a site to find these types of questions please​

Answer 1

Given:

• mass of the box =20 kg
• initial velocity pf the box = 2 m/s
• distance covered = 5 m
• final velocity = 0 m/s

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To find :

• frictional force acting on the block

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Solution:

By using the Third kinematic equation ,

>> v^2 = u^2 + 2 a s

here ,

• v = final velocity
• u = initial velocity
• a= acceleration
• s= distance

Now by substituting the given values in the above equation we get ,

-> u^2 = - 2 a s

-> a = -  u^2/ 2 s

-> a = -  4/ 2 x 5

-> a = 4 /10

-> a = 0.4 m/s^2

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frictional force acting on the box is given by the formula ,

>> F =ma

here ,

• F= frictional force
• m= mass of the box
• a= acceleration of the box

-> F= 20 x 0.4

-> F= 8 N

The frictional force acting on the box= 8 N

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Hope this helps :D