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levacccp [35]
3 years ago
14

In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw

ay. If the 3.5 kg book is pushed from rest through a distance of 0.91 m by the horizontal 25 N force from the broom and then has a speed of 1.53 m/s, what is the coefficient of kinetic friction between the book and floor?
Physics
1 answer:
RSB [31]3 years ago
5 0

Answer:

The coefficient of kinetic is

u_{k}=0.59

Explanation:

The forces in the axis 'x' and 'y' using law of Newton to find coefficient of kinetic friction

ΣF=m*a

ΣFy=W-N=0

ΣFy=Fn-Fu=m*a

F_{u} =u_{k} *N\\F_{N}=25N\\N=W\\N=3.5kg*9.8\frac{m}{s^{2} }=34.3N

F_{N}-F_{u}=m*a\\F_{N}-u_{k}*N=m*a\\u_{k}*N=F_{N}-m*a\\u_{k}=\frac{F_{N}-m*a}{N}

Now to find the coefficient can find the acceleration using equation of uniform motion accelerated

v_{f} ^{2}=v_{o}^{2}+2*a(x_{f}-x_{o})\\x_{o}=0\\v_{o}=0\\v_{f} ^{2}=2*a*x_{f}\\a=\frac{v_{f} ^{2}}{2*a*x_{f}}\\ a=\frac{(1.53\frac{m}{s} )^{2}}{2*0.91m}\\a= 1.28 \frac{m}{s^{2} }

So replacing the acceleration can fin the coefficient:

u_{k}=\frac{F_{N}-m*a }{N}\\u_{k}=\frac{25N-(3.5kg*1.28\frac{m}{s^{2}} }{34.3N} \\u_{k}=0.59

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