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The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and

Q5.36    The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

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Using Newton's second law of motion we can write :

                                                                    F\ =\ ma

                                                                            =\ 40\times 2\ =\ 80\ N

Also, the frictional force is given by :

                                                                   f\ =\ \mu mg

 or                                                                     =\ 0.15 \times 40 \times 10\ =\ 60\ N

Thus net force acting is :    80  -  60   =  20 N.

The backward acceleration is :

                                                          a_b\ =\ \frac{20}{40}\ =\ 0.5\ m/s^2

Now using the equation of motion we can write :

                                            s\ =\ ut +\ \frac{1}{2}at^2

or                                       5\ =\ 0\ +\ \frac{1}{2}(0.5)t^2

Thus                                 t\ =\ \sqrt{20}\ s

And the distance travelled by truck is :

                                         s\ =\ ut +\ \frac{1}{2}at^2

or                                           =\ 0 +\ \frac{1}{2}\times 2 (\sqrt{20})^2                             

or                                            =\ 20 \ m

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