Figure gives, as a function of time t, the force componentthat acts on a 3.00kgice block that can move only along the xaxis. At t=0 , the block is moving in the positive direction of the axis, with a speed of 3.0m/s. What are its (a) speed and (b) direction of travel at t=11 s?

It is given that:

The mass of the ice block is M = 3 kg

The initial velocity of the ice block$v_0=3m/s$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Using Newton’s second law,the acceleration and from that, the velocity of the ice block and its direction of motion can be found.

Formulae:

Newton’s second law is,

$F_{net}=\underset{}{\overset{}{\sum Ma}}$ (i)

where, $F_{net}$ is the net force, Mis mass and a is an acceleration.

The acceleration is defined as,

$$\begin{gathered} a=\frac{dv}{dt} \\ =\frac{\left(v_f-v_0\right)}{t} \end{gathered}$$ (ii)

The acceleration is defined as,

$a=\frac{dv}{dt}$

So, equation (i) can be written as,

$\frac{dv}{dt}=\frac{F_{net}}{M}$

Therefore, velocity is the integral of $\frac{F_{net}}{M}$with respect to.So, the velocity can be found from the area under the curve divided by mass. At t = 11s , the area under the curve is equal to 15 unit

Therefore, the velocity of the ice block at t = 11 sec is,

$$\begin{gathered} v-v_0=15/3 \\ =5unit \end{gathered}$$

As initial velocity of the block is 3 m/s it’s velocity at 11s is

$$\begin{gathered} v=5+3 \\ =8m/s \end{gathered}$$

As the velocity of the block at 11 is positive, the direction of travel of the block is in positive x direction.

Therefore, graph of the force vs time is equivalent to acceleration vs time if the force is divided by mass. From this graph, velocity can be found by finding the area under the curve. Using the given initial velocity, the final velocity and its direction at given time can be found.