Question

Two forces of 10 N and 30 N are applied to a 10 kg box. Find (1) the box’s acceleration when both forces point due east and (2) the box’s acceleration when 10 N force points due east and 30 N force points due west.

Answer 1

1. Answer;

4 m/s²

Explanation;

When both forces point due east then;

the resultant force will be; 10 + 30 = 40 N due east

But; From second Newton's law of motion;

F = ma

Thus;

a = F/m ; F= 40 N and mass = 10 kg

= 40/ 10

= 4 m/s² to wards east

2. Answer;

1 m/s²

Explanation;

when 10 N force points due east and 30 N force points due west then;

the resultant force will be; 30 N - 10 N = 10 N due west

But; From second Newton's law of motion;

F = ma

Thus;

a = F/m

= 10 /10

= 1 m/s² , towards the west direction

Answer 2

Step-by-step explanation:

Given that,

First force = 10 N

Second force = 30 N

Mass of box = 10 kg

(1). When both forces point due east

Then the net force will be

`F_(net)=F_(1)+F_(2)`

`F_(net)=10+30`

`F=40 N`

We need to calculate the acceleration

Using newton's second law

`F= ma`

put the value into the formula

`40=10* a`

`a=(40)/(10)`

`a=4\ m/s^2`

The box accelerate towards east.

(2). When 10 N force points due east and 30 N force points due west.

Then the net force will be

`F_(net)=F_(2)-f_(1)`

`F_(net)=30-10=20\ N`

We need to calculate the acceleration

Using newton's second law

`F= ma`

`20=10* a`

`a=(20)/(10)`

`a=2\ m/s^2`

The box accelerate towards west.

Hence, This is the required solution.