Question

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 65 Nwhen the cab is standing still. What is the reading when the cab is moving upward(a) with a constant speed of 7.6 m/sand (b) with a speed 7.6 m/s ofWhile decelerating at a rate of 2.4 m/s?

Short answer

(a) Scale reading when cab moving upward with constant speed 7.6 m/s is 65N.

(b) Scale reading when cab moving upward with speed 7.6m/s while decelerating with 2$4\mathrm{m}/{\mathrm{s}}^2$ is 49 N.

Step-by-step solution

Given information

1. T=W=65N ,
2. V=7.6m/s and
3. $a=2.4\mathrm{m}/{\mathrm{s}}^2$.

Understanding the concept

Newton’s second law states that the net force acting on the object is equal to the product of mass and the acceleration of the object.

Find the scale reading by using Newton’s law. We can find the mass of the object just using weight equation.

Formula used

1. W=mg
2. F =ma

Here, W is weight of the object,m is mass of the object,g is gravitational acceleration, F is force and is acceleration of the object.

Calculate the mass of the object

Mass of the object can be calculated using the formula for the weight.

$$\begin{gathered} W=mg \\ m=\frac{W}{g} \\ =\frac{65\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =6.6\mathrm{kg} \end{gathered}$$

Therefore, the mass of the object is 6.6kg.

(a) Calculate the scale reading when cab is moving upward with constant speed

As the cab is moving with constant speed the acceleration is zero so the scale reading will be 65N.

(b) Calculate the scale reading when cab is moving upward with speed 7.6m/s while decelerating with 2.4m/s2

We apply Newton’s second law to find tension or scale reading,

$$\begin{gathered} ma=T-mg \\ T=ma+mg \\ =m\left(a+g\right) \\ =6.6\mathrm{kg}\left(-2.4\mathrm{m}/{\mathrm{s}}^2+9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =49\mathrm{N} \end{gathered}$$

Therefore, the tension or scale reading is 49N.