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Grade: 10

                        

An object is hung from a spring scale attached to the ceiling of an elevator. The scale reads 65 N when the elevator is standing still. (a) What is the reading when the elevator is moving upward with a constant speed of 7.6 m/s? (b) What is the reading of the scale when the elevator is moving upward with a speed of 7.6 m/s and decelerating at 2.4 m/s 2 ?

5 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
							Weight of the object (W) when the elevator is standing, W = 65 N
We have to find out the scale reading when the elevator is moving upward with a constant speed of 7.6 m/s.Since the elevator is moving upward with a constant speed, therefore there is no acceleration of the system resulting there is no force. Thus the scale reading must be equal to the weight of the object and that will be 65 N.
We have to find out the scale reading when the elevator is moving upward with a speed of 7.6 m/s and decelerating at 2.4 m/s2.The force exerted on the object due to the deceleration at the rate 2.4 m/s2 (a = - 2.4 m/s2) will be,
F = ma
= (W/g) a …… (1)
Where W is the weight of the object (W = 65 N) when the elevator is at rest and g is the free fall acceleration of the object (g=9.80 m/s2).
Putting the value of W, g and a in equation (1) the force exerted on the object will be,
F = (W/g) a
= (65 N/9.80 m/s2) (-2.4 m/s2)
= -15.92 N
= -16 N …… (2) (Rounding off to two significant figure)
When the elevator is moving upward with a speed of 7.6 m/s and decelerating at 2.4 m/s2, the force would be,
= F- (-W)
= -16 N –(-65 N)
= -16 N+65 N
= 49 N …… (3)
From the above observation we conclude that, the scale reading when the elevator is moving upward with a speed of 7.6 m/s and decelerating at 2.4 m/s2 would be 49 N.

5 years ago
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