Q42P - University Physics with Modern Physics

Question

A 6.50 kg instrument is hanging by a vertical wire inside a spaceship that is blasting off from rest at the earth’s surface. This spaceship reaches an altitude of 276 m in 15.0 s with constant acceleration. (a) Draw a free-body diagram for the instrument during this time. Indicate which force is greater. (b) Find the force that the wire exerts on the instrument.

Step-by-Step Solution

Answer

(a)

The tension force is greater.

(b) The force that the wire exerts on the instrument is 79.64 N.

Step 1: Identification of the given data

The given data is listed below as:

 

• The mass of the instrument is, m =6.5 kg
• The height reached by the spaceship is, h =276 m
• The time taken by the spaceship to reach the height is, t =15 s

Step 2: Significance of the force

The force exerted by an object is directly proportional to the mass and the acceleration of that particular object. The acceleration due to gravity also depends on the force exerted on an object.

Step 3: (a) Determination of the free body diagram for the instrument and identification of force that is greater

The free-body diagram of the instrument has been drawn below:

Here, in the above diagram, the weight w of the instrument is acting downwards and the acceleration is acting upwards. Moreover, the tension T is also acting in an upward direction.

 

As the weight is acting downwards and tension is acting in the upwards direction, the instrument mainly acts in the upwards direction. Hence, the net force also acts in the upward direction which states that the tension is greater than the weight.

 

Thus, the tension force is greater.

Step 4: (b) Determination of the force exerted by the wire

The equation of the acceleration of the wire is expressed as:

$$\begin{gathered} h=ut+\frac{1}{2}a{t}^2 \\ h-ut=\frac{1}{2}a{t}^2 \\ 2\left(h-ut\right)=a{t}^2 \\ a=\frac{2\left(h-ut\right)}{t^2} \end{gathered}$$ 

 

Here, a is the acceleration of the wire, h is the height reached by the spaceship, u is the initial speed of the instrument and t is the time taken by the spaceship to reach the height.

 

As the instrument was at rest initially, then the initial speed of the instrument is zero.

 

The equation of the force exerted by the wire is expressed as:

$$\begin{gathered} F-mg=ma \\ F=mg+ma \\ =m\left(g+\frac{2\left(h-ut\right)}{t^2}\right) \end{gathered}$$ 

 

Here, F is the force exerted by the wire and m is the mass of the instrument.

 

Substitute all the values in the above equation.

 

$$\begin{gathered} \mathrm{F}=\left(6.5 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2+\frac{2\left(276 \mathrm{m}-\left(0\right)\mathrm{t}\right)}{{\left(15 \mathrm{s}\right)}^2}\right) \\ =\left(6.5 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2+\frac{\left(552 \mathrm{m}\right)}{\left(225 {\mathrm{s}}^2\right)}\right) \\ =\left(6.5 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2+2.45 \mathrm{m}/{\mathrm{s}}^2\right) \\ =\left(6.5 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2+2.45\mathrm{m}/{\mathrm{s}}^2\right) \\ =\left(6.5 \mathrm{kg}\right)\left(12.25 \mathrm{m}/{\mathrm{s}}^2\right) \end{gathered}$$

 

Hence, further as:

$$\begin{gathered} \mathrm{F}=79.64 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =79.64 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1 \mathrm{N}}{1 \mathrm{kg}./{\mathrm{s}}^2}\right) \\ =79.64 \mathrm{N} \end{gathered}$$ 

 

Thus, the force that the wire exerts on the instrument is 79.64 N.