Respuesta :
Answer:
Explanation: Scale only reads reaction force so we will find out reaction force of ground in different cases.
a ) Let R be the reaction force of floor in the first case.
Net force = R - mg
= mass x acceleration = 0 ( body is at rest ).
R = mg , scale will read his actual weight.
In the case of b) and c) also acceleration is zero hence
R - mg =0
R =mg ,scale will read his actual weight.
d) In this case there is acceleration of 3 m s⁻² in upward direction so
R - mg = ma
R = m( g+a) = 75 x ( 9.8 + 3 ) = 960 N
The reading of scale will be more than his actual weight .
It will read weight as 960/9.8 = 97.96 kg.
e) In this case body is going downwards
mg -R = ma
R = m ( g - a )
= 75 ( 9.8 - 3 )
= 435 N
= 435 / 9.8 kg
= 44.38 kg
So scale will read less than his actual weight.
Scale reading of the person weight (75 kg) change with different rate of speed of elevator due to acceleration.
- (a) Scale reading when the elevator is at rest is 75 kg.
- (b) Scale reading when the elevator is ascending at a constant speed of 3.0m/s is 75 kg.
- (c) Scale reading when the elevator is falling at 3.0m/s is 75 kg.
- (d)Scale reading when the elevator is accelerating upward at 3.0m/s2 is 97.96kg.
- (e)Scale reading when the elevator is accelerating downward at 3.0m/s2 is 44.38 kg.
What is effect of acceleration on the scale reading?
The reaction force applied on the measurement scale changes its reading with different values of acceleration of object, as the reaction force of object is result of mass times acceleration of object.
Given information-
The weight of the person is 75 kg.
- (a)Scale reading when the elevator is at rest-
In case of the rest, the total force actin on the elevator is due to the mass of the person and the gravity of the Earth.
Thus, scale reading when the elevator is at rest is,
[tex]r=mg\\r=W\\r=75\rm kg[/tex]
Thus, scale reading when the elevator is at rest is 75 kg.
- (b) Scale reading when the elevator is ascending at a constant speed of 3.0m/s,
The speed of the elevator is constant. Thus the rate of change of velocity is also constant. For this case the acceleration of elevator is zero.
Thus net force acting on the elevator is,
[tex]R=0+mg\\R=w\\R=75\rm kg[/tex]
Hence, the scale reading when the elevator is ascending at a constant speed of 3.0m/s is 75 kg.
- (c) Scale reading when the elevator is falling at 3.0m/s,
The speed of the elevator is constant , which is 3 m/s. Thus the rate of change of velocity is also constant. For this case the acceleration of elevator is zero.
Thus net force acting on the elevator is,
[tex]R=0+mg\\R=w\\R=75\rm kg[/tex]
Hence, the scale reading when the elevator is falling at 3.0m/s is 75 kg.
- (d)Scale reading when the elevator is accelerating upward at 3.0m/s2
The acceleration is is in upward direction. Thus the net acceleration effect is due to the sum of gravity and the acceleration of 3 m/s. Thus the scale reading of the elevator is,
[tex]r=\dfrac{75}{9.81} )(9.81+3)\\r=97.96\rm kg[/tex]
Thus, scale reading when the elevator is accelerating upward at 3.0m/s2 is 97.96 kg.
- (e)Scale reading when the elevator is accelerating downward at 3.0m/s2 .
The acceleration is is in downward direction. Thus the net acceleration effect is due to the difference of gravity and the acceleration of 3 m/s. Thus the scale reading of the elevator is,
[tex]r=\dfrac{75}{9.81} )(9.81-3)\\r=44.38\rm kg[/tex]
Thus, scale reading when the elevator is accelerating downward at 3.0m/s2 is 44.38 kg.
Hence,
- (a) Scale reading when the elevator is at rest is 75 kg.
- (b) Scale reading when the elevator is ascending at a constant speed of 3.0m/s is 75 kg.
- (c) Scale reading when the elevator is falling at 3.0m/s is 75 kg.
- (d)Scale reading when the elevator is accelerating upward at 3.0m/s2 is 97.96kg.
- (e)Scale reading when the elevator is accelerating downward at 3.0m/s2 is 44.38 kg.
Learn more about the acceleration effect on measurement here;
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