Question

A force of 40 N is applied to a 6 kg box of books. If the frictional force acting on the box is 16 N, and the box was initially at rest, what is the velocity of the box 10 s later? What is the box’s displacement over the 10 s period?

Answer 1

The final velocity is 40 m/s. The box will undergo a displacement of 200 m over the period of 10 s.

Answer:

Step-by-step explanation:

So the net force acting on the box of books will be the difference of normal force and frictional force acting on the box. As the normal force acting on the box is 40 N and the frictional force is 16 N, then the net force is

Net force = 40 N - 16 N = 24 N.

Also, the mass of the box is given as 6 kg. So as per second law of motion, the acceleration can be found by ratio of net force to mass of the object.

Acceleration of the box = Net force/Mass = 24/6=4 m/s².

Then velocity can be determined using the first equation of motion.

v = u + at =0 + (4× 10 s) = 40 m/s.

So the final velocity is 40 m/s.

Similarly, by using the second equation of motion,

s = ut +1/2 at²

s = (0×10)+(0.5×4×10×10)=200 m

And so, the box will undergo a displacement of 200 m over the period of 10 s.