Taking g=9.81m/s^2
The astronaut will experience normal weight (due to gravity) plus weight due to acceleration.
Apparent weight = 95kg * (1+32.0/9.81)
The force experienced is: 95 *(9.81+32.0) N
Suppose a rocket launches with an acceleration of 32.0m/s^2.
What is the apparent weight of an 95-kg astronaut aboard this rocket?
2 answers
Graham is correct; I just do not like the way that it is explained in some examples solved. I saw this earlier, but left it alone, and went and did something else.
The resultant force points up and you have the force of gravity point down subtracting from the weight. So the resultant force is the weight of the astronaut minus the the gravitational force.
ΣF=ma=W-Fg
ma=W-Fg
(95kg)*(32.0m/s^2)=Fw-95kg*(9.8m/s^2)
Solving for Fw,
Fw=(95kg)*(32.0m/s^2)+(95kg)*(9.8m/s^2)
Fw=95kg(32.0m/s^2+9.8m/s^2)
Solve
Graham and I are in agreement; I just explained differently.
The resultant force points up and you have the force of gravity point down subtracting from the weight. So the resultant force is the weight of the astronaut minus the the gravitational force.
ΣF=ma=W-Fg
ma=W-Fg
(95kg)*(32.0m/s^2)=Fw-95kg*(9.8m/s^2)
Solving for Fw,
Fw=(95kg)*(32.0m/s^2)+(95kg)*(9.8m/s^2)
Fw=95kg(32.0m/s^2+9.8m/s^2)
Solve
Graham and I are in agreement; I just explained differently.
1 answer