Asked by Mariah
A contestant in a winter games event pushes a 59.0 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. The block is being pushed at a direction of 23 degrees below the horizontal.
(a) Calculate the minimum force F he must exert to get the block moving.
(b) What is its acceleration once it starts to move, if that force is maintained?
I am stuck when it comes to the x and y direction and how they are related. Please help!
(a) Calculate the minimum force F he must exert to get the block moving.
(b) What is its acceleration once it starts to move, if that force is maintained?
I am stuck when it comes to the x and y direction and how they are related. Please help!
Answers
Answered by
Anonymous
(a)
Fnet = ma = 0
Fnet = Fapp - Ff
Fapp - Ff = 0
Fapp = Ff
F = (Fn)(μs)
F = (Fg)(0.1)
F = (9.8m)(0.1)
F = (9.8)(59)(0.1)
F = 57.82 N
(b)
Fnet = Fapp - Ff
F = 57.82 - (Fn)(μk)
F = 57.82 - (Fg)(0.03)
F = 57.82 - (9.8m)(0.03)
F = 57.82 - (9.8)(59)(0.03)
F = 57.82 - 17.346 = 40.474
F = ma = 40.474
(59)a = 40.474
a = 0.686 m/s²
The x and y directions are related because friction (a horizontal force that pushes against your push) is based on the normal force (which in this case is vertical, since the lake is a flat surface and the only force pushing down on the block is gravity).
Fnet = ma = 0
Fnet = Fapp - Ff
Fapp - Ff = 0
Fapp = Ff
F = (Fn)(μs)
F = (Fg)(0.1)
F = (9.8m)(0.1)
F = (9.8)(59)(0.1)
F = 57.82 N
(b)
Fnet = Fapp - Ff
F = 57.82 - (Fn)(μk)
F = 57.82 - (Fg)(0.03)
F = 57.82 - (9.8m)(0.03)
F = 57.82 - (9.8)(59)(0.03)
F = 57.82 - 17.346 = 40.474
F = ma = 40.474
(59)a = 40.474
a = 0.686 m/s²
The x and y directions are related because friction (a horizontal force that pushes against your push) is based on the normal force (which in this case is vertical, since the lake is a flat surface and the only force pushing down on the block is gravity).
Answered by
msa
This is wrong
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