Question

A body of mass 2kg was moving with a velocity of 10 ms-1 when a force F is applied on it. Find the magnitude and direction of F if (1) the body stops after 10s. (2) it acquires a velocity of 20 ms-1 in travelling a distance of 15m after application of the force.

Answer 1

1)

Given
initial velocity (u) = 10 m/s
Final velocity (v) = 0 m/s (since the body stops)
time = 10 s


So acceleration =
$\frac{v - u}{t} \\ \\ = > \frac{0 - 10}{10} \\ \\ = > - 1$

So acceleration = -1 m/s²

Now,

F = ma

=> F = 2 × -1

=> F = -2 N

(Negative sign means direction is opposite to the motion)

Hence magnitude of F = 2 N
Direction = Opposite to the motion happening earlier.

2)
Given
initial velocity (u) = 10 m/s
Final velocity (v) = 20 m/s
displacement = 15 m

By applying the third equation of motion :-
${v}^{2} - {u}^{2} = 2as \\ \\ = > {20}^{2} - {10}^{2} = 2a \times 15 \\ \\ = > 400 - 100 = 30a \\ \\ = > 30a = 300 \\ \\ = > a = \frac{300}{30} \\ \\ = > a = 10$

So acceleration = 10 m/s²

Now,

F = ma

=> F = 2 × 10

=> F = 20 N

Since the force is positive, the Direction is same as the motion

=> Magnitude = 20 N
Direction = same as motion


Hope it helps dear friend ☺️✌️