A contestant in a winter games event pushes a 41.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in the figure. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
(a) Calculate the minimum force F (in N) he must exert to get the block moving.
(b) What is its acceleration (in m/s2) once it starts to move, if that force is maintained?
F\' 1 25°
Solution
a) Apply, Fnety = 0
N + F*sin(25) - m*g = 0
N = m*g - F*sin(25)
Let F the minimum force he must exert to get the block moving
so,
F*cos(25) = fs_max
= mue_s*N
F*cos(25) = mue_s*(m*g - F*sin(25))
F*cos(25) = mue_s*m*g - mue_s*F*sin(25)
F*(cos(25) + mue_s*sin(25)) = mue_s*m*g
==> F = mue_s*m*g/(cos(25) + mue_s*sin(25))
= 0.1*41*9.8/(cos(25) + 0.1*sin(25))
= 42.3 N <<<<<<<<------------------------------Answer
b)
let a is the accleration of the object.
Normal force on the object,
N = m*g - F*sin(25)
= 41*9.8 - 42.3*sin(25)
= 384 N
Fnetx = F*cos(25) - fk
m*a = F*cos(25) - mue_k*N
a = (F*cos(25) - mue_k*N)/m
= (42.3*cos(25) - 0.03*383)/41
= 0.655 m/s^2Â Â <<<<<<<<------------------------------Answer
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A contestant in a winter games event pushes a 41-0 kg block of ice acr.docx
1. A contestant in a winter games event pushes a 41.0 kg block of ice across a frozen lake with a
rope over his shoulder as shown in the figure. The coefficient of static friction is 0.1 and the
coefficient of kinetic friction is 0.03.
(a) Calculate the minimum force F (in N) he must exert to get the block moving.
(b) What is its acceleration (in m/s2) once it starts to move, if that force is maintained?
F' 1 25°
Solution
a) Apply, Fnety = 0
N + F*sin(25) - m*g = 0
N = m*g - F*sin(25)
Let F the minimum force he must exert to get the block moving
so,
F*cos(25) = fs_max
= mue_s*N
F*cos(25) = mue_s*(m*g - F*sin(25))
F*cos(25) = mue_s*m*g - mue_s*F*sin(25)
F*(cos(25) + mue_s*sin(25)) = mue_s*m*g
==> F = mue_s*m*g/(cos(25) + mue_s*sin(25))
= 0.1*41*9.8/(cos(25) + 0.1*sin(25))
= 42.3 N <<<<<<<<------------------------------Answer
2. b)
let a is the accleration of the object.
Normal force on the object,
N = m*g - F*sin(25)
= 41*9.8 - 42.3*sin(25)
= 384 N
Fnetx = F*cos(25) - fk
m*a = F*cos(25) - mue_k*N
a = (F*cos(25) - mue_k*N)/m
= (42.3*cos(25) - 0.03*383)/41
= 0.655 m/s^2Â Â <<<<<<<<------------------------------Answer
