91影视

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鈥檚 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鈥檚 second law,the acceleration and from that, the velocity of the ice block and its direction of motion can be found.

Formulae:

Newton鈥檚 second law is,

$F_{net}=\underset{}{\overset{}{鈭�Ma}}$ (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鈥檚 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.