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A contestant in a winter games event pushes a 51.0 kg block of ice across a frozen lake as shown in the figure below. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03. (Assume ? = 20Â°.) a) Calculate the minimum force F (in N) he must exert to get the block moving. b) What is its acceleration (in m/s 2 ) once it starts to move, if that force is maintained? Can someone please help me with part b? Thank you! A contestant in a winter games event pushes a 51.0 kg block of ice across a frozen lake as shown in the figure below. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03. (Assume e 20Â°.) (a) Calculate the minimum force F (in N) he must exert to get the block moving. 55.2 What is its acceleration (in m/s2) once it starts to move, if that force is maintained? 2 0.78 (b) Consider Newton\'s Second Law. m/s Solution Let the force is F Fcos(theta)=us*N Fsin(theta)+N=mg N=mg-Fsin(20)= Fcos(20)=0.1*(51*9.8-Fsin(20)) F=51.3 N let acceleration is a 51.3*cos(20)-0.03*(51*9.8-51.3sin(20))=51a a=0.67 m/s^2 .

A contestant in a winter games event pushes a 51.0 kg block of ice across a frozen lake as shown
in the figure below. The coefficient of static friction is 0.1 and the coefficient of kinetic friction
is 0.03. (Assume ? = 20Â°.)
a) Calculate the minimum force F (in N) he must exert to get the block moving.
b) What is its acceleration (in m/s 2
) once it starts to move, if that force is maintained?
Can someone please help me with part b? Thank you!
A contestant in a winter games event pushes a 51.0 kg block of ice across a frozen lake as shown
in the figure below. The coefficient of static friction is 0.1 and the coefficient of kinetic friction
is 0.03. (Assume e 20Â°.) (a) Calculate the minimum force F (in N) he must exert to get the
block moving. 55.2 What is its acceleration (in m/s2) once it starts to move, if that force is
maintained? 2 0.78 (b) Consider Newton's Second Law. m/s
Solution
Let the force is F
Fcos(theta)=us*N
Fsin(theta)+N=mg
N=mg-Fsin(20)=
Fcos(20)=0.1*(51*9.8-Fsin(20))
F=51.3 N
let acceleration is a
51.3*cos(20)-0.03*(51*9.8-51.3sin(20))=51a
a=0.67 m/s^2

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