Question

A student’s backpack, full of textbooks, is hung from a spring scale attached to the ceiling of an elevator. When the elevator is accelerating downward at 3.8m/s2, the scale reads 60N. a) What is the mass of the backpack? b) What does the scale read if the elevator moves upward while slowing down at a rate 3.8 m/s2? c) What does the scale read if the elevator moves upward at constant velocity? d) If the elevator had no brakes and the cable supporting it were to break loose so that the elevator could fall freely, what would the spring scale read? 6

Answer 1

a) 10 kg

b) 60 N

c) 98 N

d) 0 N

Explanation:

a)

In this problem, there are two forces acting on the backpack:

- The restoring force from the spring, upward, of magnitude $F=60 N$

- The weight of the backpack, downward, of magnitude $mg$, where

m = mass of the backpack

$g=9.8 m/s^2$ acceleration due to gravity

So the net force on the backpack is (taking upward as positive direction)

$F_{net}=F-mg$

According to Newton's second law of motion, the net force must be equal to the product between the mass of the backpack and its acceleration, so

$F-mg=ma$

where

$a=-3.8 m/s^2$ is the acceleration of the backpack and the elevator, downward (so, negative)

If we solve the formula for m, we can find the mass of the backpack:

$F=m(a+g)\\m=\frac{F}{a+g}=\frac{60}{-3.8+9.8}=10 kg$

b)

In this case, the elevator is moving upward, and it is slowing down at a rate of $3.8 m/s^2$.

Since the elevator is slowing down, it means that the direction of the acceleration is opposite to the direction of motion: and since the elevator is moving upward, this means that the direction of the acceleration is downward: so the acceleration is negative,

$a=-3.8 m/s^2$

The net force acting on the backpack is still:

$F_{net}=F-mg$

where

F is the restoring force in the spring, which this time is unknown (it corresponds to the reading on the scale)

Using again Newton's second law of motion,

$F-mg=ma$

Therefore in this case, the reading on the scale will be:

$F=m(g+a)=(10)(9.8-3.8)=60 N$

So the reading is the same as in part a).

c)

In this case, the elevator is moving at a constant velocity.

The net force on the backpack is still:

$F_{net}=F-mg$

However, since here the elevator is moving at constant velocity, and acceleration is the rate of change of velocity, this means that the acceleration of the elevator is zero:

$a=0$

So Newton's second law of motion can be written as:

$F_{net}=0$

So

$F-mg=0$

Which means that the reading on the scale is equal to the weight of the backpack:

$F=mg=(10)(9.8)=98 N$

d)

In this  case, the elevator had no brakes and the cable supporting it breaks loose.

This means that the elevator is now in free fall.

So its acceleration is simply the acceleration due to gravity (which is the acceleration of an object in free fall):

$a=-9.8 m/s^2$

And the direction is downward, so it has a negative  sign.

The net force on the backpack is still

$F_{net}=F-mg$

So Newton's second law can be rewritten as

$F-mg=ma$

Therefore, we can re-arrange the equation to find F, the reading on the scale, and we find:

$F=m(g+a)=(10)(9.8-9.8)=0 N$

So, the reading on the scale is 0 N.