Question

a constant force of 12n is exerted for 8seconds on a 24kg object initially at rest change of speed of this object be

Answer 1

Here,

F = 12N

M = 24Kg

t = 8s

initial velocity = 0 m/s

final velocity = ?

So, we know that F = ma

12 = 24a

a = 12/24

a = 0.5 meter per second sq.

By applying 1st equation of motion,

v = u+at

v= 0+0.5x8

v=4 m/s

Answer 2

The  change of speed of this object be 4 meter/second.

Explanation:

Given:

A constant force of  12 n is exerted for 8 seconds on a 24 kg object initially at rest.

To Find:

The  change of speed of this object.

Solution:

As given, a constant force of 12 n is exerted for 8 seconds on a 24 kg object initially at rest.

Initial velocity of object, u = 0

Final velocity of object,v = ?

Force, F = 12 N

Time,t = 8 seconds

Mass of object, m = 24 Kg

Using newton second law.

$F=\frac{m(v-u)}{t}$

$12=\frac{24(v-0)}{8}$

$12=\frac{24(v)}{8}$

$12=3v$

$v=4\ meter/second$

Therefore, change of speed of this object = v-u

                                                                       $=4-0$

                                                                        $=4 \ meter/second$

Thus,the  change of speed of this object be 4 meter/second.

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