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We have that for the Question " constant applied force Fp of 11.0 N pushes a box with a mass=7 kg a distance x=15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is .110." it can be said that
the final velocity vf is
[tex]vf=3.83m/s[/tex]
and the applied force Fnew is
[tex]F_{new}=3.43N[/tex]
From the question we are told
A constant applied force Fp of 11.0 N pushes a box with a mass=7 kg a distance x=15.0 m across a level floor. The coefficient of kinetic friction between the box and the floor is .110. Assuming the box starts from the rest, what is the final velocity vf of the box at 15.0 m point? If there were no friction between the box and the floor, what applied force F new would give the box the same final velocity?
Generally the equation for the Force is mathematically given as
F_p-\mu mg=ma
Therefore
11-0.110*7*9.8=7*a
a=0.49m/s^2
Using Newtons equation
v^2=u^2+2as
v^2=0+2*0.49*15
v=3.83m/s
Therefore
F=ma
F=0.49*7
F_{new}=3.43N
Therefore
the final velocity vf is
[tex]vf=3.83m/s[/tex]
and the applied force Fnew is
[tex]F_{new}=3.43N[/tex]
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