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**Hint**: In question A, we will first find out the acceleration using Newton’s second law of motion, which states that the force applied on or by a body is the product of the mass and the acceleration of the body due to that force.

Then using that, we will find out the time using the equations of motion.

In question B, we will first find out the acceleration using the equations of uniform motion and then find out the force using Newton’s second law of motion.

Formula used:

$t=\dfrac{v-u}{a}$

\[F=ma\]

$a=\dfrac{v-u}{t}$

Where F is the force, m is the mass of the body, a is the acceleration, t is the time interval and v, u are the final and initial velocities respectively.

Complete step by step answer:

First, we will identify all the information given to us.

Force applied on body (F) = -50N (negative since it is a retarding force and retarding forces are negative)

Mass of body (m) = 20 Kg

initial velocity (u) = $15m{{s}^{-1}}$

Final velocity (v) = $0m{{s}^{-1}}$ (since the body comes to a stop)

Therefore, using Newton’s second law of motion which states that the force applied on or by a body is the product of the mass and the acceleration of the body due to that force, that is, \[F=ma\], we get,

$-50=20\times a$

$\therefore a=\dfrac{-50}{20}=-2.5m{{s}^{-2}}$ ---(1)

Now for finding out time taken for the body to stop, we will use the equation of uniform motion, $t=\dfrac{v-u}{a}$.

$t=\dfrac{0-15}{-2.5}=6s$

Therefore the required time taken is 6 seconds.

B. First, we will identify all the information given to us.

Mass of body (m) = 3 Kg

initial velocity (u) = $2m{{s}^{-1}}$

Final velocity (v) = $3.5m{{s}^{-1}}$

Time taken by the body (t) = 25s

Now, using the equation of uniform motion, we will find out the acceleration (a). Using $a=\dfrac{v-u}{t}$

$a=\dfrac{3.5-2}{25}=\dfrac{1.5}{25}=0.06m{{s}^{-2}}$ --(1)

Now, using Newton’s second law of motion, which states that the force applied on or by a body is the product of the mass and the acceleration of the body due to that force, that is, \[F=ma\],

Using (1) , we get

$F=3\times 0.06=0.18N$

Hence, the required magnitude of force is 0.18N.

**Note**: Students must take care of the sign of the acceleration and the force while writing the equations since a negative sign implies that the force is opposing and tries to resist the motion of the body. Simply writing the magnitude of the force without factoring in the direction will lead to an erroneous result. One must also remember that the direction of the net force on a body must be equal to the direction of the acceleration of the body.

Students must check whether the acceleration of the body in the question is constant or not, since only then equations of motion for constant acceleration can be applied. If the force on the body and hence, the acceleration is different in different parts of the problem, then the problem should be broken down into that many parts and each part solved individually for which the acceleration of the body remains constant.

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