Question

A block of mass 5 kg is moving horizontally at a speed of 1.5 ms^-1. A vertically upward force 5 N acts on it for 4 seconds. What will be the distance of the block from the point where the force starts

acting?

(a) 2 m (b) 6 m (c) 8 m (d) 10 m​

Answer 1

Answer:

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Step-by-step explanation:

Assume initial velocity of 1.5m/s is in the x−direction

Since there are no forces on it in this direction, there will be no acceleration. So distance =s(x)=1.5m/s(4sec)=6m

In the y−direction,F=5N and since m=5kg, Newton's 2nd Law tells us acceleration a(y)=F/m=5/5=1N/kg=1ms

2

s(y)=12a(y)t

2

=12(1)(42)=8m

Resolving the x and y vector, we note the Pythagorean Triple of 6,8,10.

So, the distance traveled is 10m

Hence,

option (D) is correct answer.

Answer 2

*Given:-

m = 5 kg

v = 1.5 m/s

F = 5 N

T = 4 sec

*Solution:-

→ Distance covered in x Distance

     ⇒ $S_n$ = 1.5 * 4

               = 6m.

_-_-_-_-_-_-__---_-_-_-_-__-.

∴ Accleration in direction:

 ⇒ $S_y = \frac{F}{m}$

⇒    $= \frac{5}{5} = 1m/s^2$

∴ distance in y direction .

⇒ $S_y = \frac{1}{2} * ayt^2$

⇒ $S_y = \frac{1}{2} * 1 * 4^2 = 8m$

_-_-_-_-_-_-_-_-_-_-_-_-_-_-..

∴ $S = \sqrt{(S_x)^2 + (S_y)^2}$

 $S = \sqrt{36+64}$

 $S = \sqrt{100}$

 $S = 10 m$

_-_-_-_-_-_-_-_-_--_-_--_-_-_-_.

option (D)

Thank you......