Question

You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain tool is pushed on a frictionless horizontal surface by a 12.8 N force, it moves 17.0 m in the first 2.10 s, starting from rest. You next observe that if you release this tool from rest at 10.0 m above the ground, it takes 2.88 s to reach the ground.What does the tool weigh on Newtonia (in N)?What would it weigh on Earth (in N)?

Answer 1

Explanation:

The given data is as follows.

    $v_{o}$ = 0,     $x_{o}$ = 0,      $t_{o}$ = 0

    $x_{1}$ = 17.0 m,      $t_{1}$ = 2.10 sec

As the force P is constant and the mass "m" of the tool is constant then it means that the acceleration "a" will also be constant.

Now,

      $x_{1} = x_{o} v_{o}t_{1} + \frac{1}{2}at^{2}_{1}$

                 = $0 + (0)(2.10 s) + \frac{1}{2}a(2.10)^{2}$

               17.0 = 2.205a

               a = $7.71 m/s^{2}$

Also, we know that

                       F = $\frac{m}{a}$

                   m = $\frac{F}{a}$

So,              m = $\frac{12.8 N}{7.71 m/s^{2}}$

                       = 1.66 kg

Since, the tool is subject to its weight W and is in free fall. Hence,

       $x_{1} = x_{o} + v_{o}t_{1} + \frac{1}{2}gt^{2}_{1}$

     10.0 m = $0 + (0)(2.88 s) + \frac{1}{2} \times g \times (2.88s)^{2}$        

           g = 2.411 $m/s^{2}$

Hence, weight of tool in Newtonia is as follows.

                   W = mg

                       = $1.66 kg \times 2.411 m/s^{2}$

                       = 4.00 N

Hence, weight of the tool on Newtonia is 4.00 N.

And, weight of the tool on the Earth is as follows.

                  W = $2.411 \times 9.80$

                      = 23.62 N

Hence, weight of the tool on Earth is 23.62 N.