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**Question**

The 0.8 kg physics book in the figure below is connected by a string to a 400 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are µ_{s} = 0.50 and µ_{k} = 0.20.

Solution

5 (1 Ratings )

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**Question**

The 0.8 kg physics book in the figure below is connected by a string to a 400 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are µ_{s} = 0.50 and µ_{k} = 0.20.

** **

**Solution**

net force acting on the book is Fnet = -mB*g*sin(20)-fk-T = mB*a

net force ating on cup is T-mc*g = mc*a

adding these two equations

-mB*g*sin(20)-mu_k*mB*g*cos(20)-mc*g = (mB+mC)*a

acclearation a = -[mB(mu_k*cos(20)+sin(20))+mc]*g/(mB+mC)

a = -[(0.8*(0.2*cos(20)+sin(20))) + 0.4]*9.81/(0.8+0.4)

a = -6.53 m/s^2

apply v^2-u^2 = 2*a*s

0^2-3^2 = -2*6.53*S

S= 0.689 m

-------------------------------------

maxmim static frictional force fs_max = mu_s*mB*g*cos(20) = 0.2*0.8*9.81*cos(20) = 1.5 N

but when the book reaches maximum height

T = mc*g = 0.4*9.81 = 3.924 N

net force on book is Fnet = fs-T-mB*g*sin(20) = 0

fs = (3.924)+(0.8*9.81*sin(20)) = 6.34 N

since fs>fs_max,the book slide back down

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