Question

1)1 kg box is pulled along a horizontal surface by a force F, so the box is moving at a constant velocity. If the coefficient kinetic friction is 0.1, determine the magnitude of the force F! (g = 9.8 m/s2)

Answer 1

Explanation:

Here,

Mass of the block,m=5kg

Coefficient of friction,μ=0.2

Distance traveled,d=20m

Let the force be F.F makes an angle 45

0

with the horizontal.

The vertical component of the applied force is Fsin45.

Thus the net force by the block on the ground =(mg–Fsin45)

So, the frictional force acting on the block is =μ(mg–Fsin45)

The block moves at a constant speed. That means the horizontal component of the force just nullifies the frictional force.

So,Fcos45=μ(mg–Fsin45)

=>Fcos45+μFsin45=μmg

=>(Fcos45)(1+μtan45)=μmg

=>(Fcos45)(1+μ)=μmg

=>Fcos45=μmg/(1+μ)

Now the work done on the block is $4= (F)(d)(cos45)$$

=(Fcos45)(d)

=[μmg/(1+μ)](d)

=(0.2)(5)(9.8)(20)/(1+0.2)

=196/1.2

=163.33J

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