Asked by Amelia
An inventive child named Chris wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley, Chris pulls on the loose end of the rope with such a force that the spring scale reads 265 N. Chris’s true weight is 287 N, and the chair weighs 223 N.
a) Find the magnitude of the acceleration of the system. The acceleration due to gravity is g = 9.8 m/s2 .
Answer in units of m/s2
b) Find the magnitude of the force Chris exerts on the chair.
Answer in units of N
a) Find the magnitude of the acceleration of the system. The acceleration due to gravity is g = 9.8 m/s2 .
Answer in units of m/s2
b) Find the magnitude of the force Chris exerts on the chair.
Answer in units of N
Answers
Answered by
Elena
The rope it attached to the man’s hands, and the chair he is in. So,
there are two tensions pulling upward on the “child-chair” system.
m=(287+223)/g=510/g,
T=265 N
2Tmg ma
ma= ( 2T-mg)/m =
=g (2•265- 510)/510=
=9.8(530-510)/510=
=0.38 m/s²
N=mg+ma = 287 + 287•0.38/9.8 = 298.13 N
there are two tensions pulling upward on the “child-chair” system.
m=(287+223)/g=510/g,
T=265 N
2Tmg ma
ma= ( 2T-mg)/m =
=g (2•265- 510)/510=
=9.8(530-510)/510=
=0.38 m/s²
N=mg+ma = 287 + 287•0.38/9.8 = 298.13 N
Answered by
Leah
Elena is correct for part a but for part b you need to subtract the tension for the answer.
N=mg+ma-T = 287 + 287•0.38/9.8 -265 = 33.13 N
N=mg+ma-T = 287 + 287•0.38/9.8 -265 = 33.13 N
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