jack gives 50 kg jill a ride on a sleigh for 500 meters. he exerts a force of 45 Newtons on the rope of the sled at an angle of 30 degrees. He pulls her at a constant velocity. What is the coefficient of friction between the sled and the ground? How much work does he do on the sled?
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ok, im not good at physics but i'll give it a try. this may be a silly question but the ground they are on is snow right?
yes i guess it would be snow
The equation of motion is (in vector form)
vector (m•g) + vector F(fr)+ vector N+vectorF =0
Projections on the axes
x: F•cosα-F(fr) = 0
y: -mg+N+F•sinα = 0
F(fr) = F•cosα,
N =mg- F•sinα,
F(fr) = k•N =k•(mg- F•sinα),
F•cosα= k•(mg- F•sinα),
k = F•cosα/(mg- F•sinα) =
= 45•0.866/(50•9.8-45•0.5) = 0.083.
W= W(F) + W(fr)=F•cosα•s +F(fr) •s=
= F•cosα•s +k•N•s =
= s• (F•cosα + k•(mg - F•sinα)) =
=500 (45•0.866 + 0.083•(50•9.8 - 45•0.5)=38885 J
vector (m•g) + vector F(fr)+ vector N+vectorF =0
Projections on the axes
x: F•cosα-F(fr) = 0
y: -mg+N+F•sinα = 0
F(fr) = F•cosα,
N =mg- F•sinα,
F(fr) = k•N =k•(mg- F•sinα),
F•cosα= k•(mg- F•sinα),
k = F•cosα/(mg- F•sinα) =
= 45•0.866/(50•9.8-45•0.5) = 0.083.
W= W(F) + W(fr)=F•cosα•s +F(fr) •s=
= F•cosα•s +k•N•s =
= s• (F•cosα + k•(mg - F•sinα)) =
=500 (45•0.866 + 0.083•(50•9.8 - 45•0.5)=38885 J
1 answer