Question

A body of mass 10 kg is moving with a velocity of 1 ms^-1. Find (i) the magnitude of the force required to stop the body
(ii) the distance which the body moves through which the body move through before coming
to rest.​

Answer 1

Answer:

$\huge{\underline{\pink{\tt{Given,}}}}$

A stone is dropped from the top of a 40 m(height) high tower.

Time(t) = 2 sec.

Initial Velocity(u) = 0

Gravitational Acceleration(g) = 9.8 m/s²

$\huge{\underline{\pink{\tt{To\:Find,}}}}$

Find the speed with which the stone strikes the ground.

$\huge{\underline{\pink{\tt{Formula\:Used:}}}}$

$\bigstar \boxed{\sf{v = u + gt}}$

$\bigstar \boxed{\sf{v^{2} = u^{2} + 2gh}}$

$\huge{\underline{\pink{\tt{Solution :}}}}$

According to the First Condition :-

Speed After 2 Seconds.

$\longrightarrow \underline{\sf{We\:know\:that,}}$

$\bigstar \boxed{\sf{v = u + gt}}$

Now, Put the Value in this Formula :-

$\longmapsto \sf{v = 0 + 9.8 \times 2}$

$\longmapsto \boxed{\sf{v = 19.6\:m/s}}$

Therefore,

The Speed of stone after 2 second will be 19.6 m/s.

$\underline{\red{\sf{Now,We\:will\:find\:the\:speed\:with\:which\:the\:stone\:strikes\:the\:ground :-}}}$

∴We know that :-

$\bigstar \boxed{\sf{v^{2} = u^{2} + gh}}$

$\longmapsto \sf{v^{2} = 0 + 2 \times 9.8 \times 40}$

$\longmapsto \sf{v^{2} = 784}$

$\longmapsto \sf{v = \sqrt{784}}$

$\longmapsto \boxed{\sf{v = 28\:m/s}}$

Therefore,

Speed with which the stone strikes the ground is 28 m/s.

$\rule{200}2$