Answer to Question #163890 in Mechanics | Relativity for Emma Lidou

Question

A 40-kg trunk sliding forward across a floor slows down from 5.0 to 2.0 m/s in 6.0 s. Assuming that the force acting on the trunk is constant, what is its magnitude and direction?

Answer 1

Let's first find the acceleration of the trunk:

$$a=\dfrac{v-v_0}{t}=\dfrac{2.0\ \dfrac{m}{s}-5.0\ \dfrac{m}{s}}{6.0\ s}=-0.5\ \dfrac{m}{s^2}.$$

The sign minus means that the trunk is decelerates.

Finally we can find the force that acts on the trunk:

$$F=ma=40\ kg\cdot(-0.5\ \dfrac{m}{s^2})=-20\ N.$$

The magnitude of the force acting on the trunk equals 20 N. The sign minus means that the force directed in the opposite direction (backward) to the motion of the trunk.

Answer:

20 N, backward.