Question

a mass of 10 g m is moving with an acceleration of 5cm per second square what is the force acting on it​

Answer 1

Given:-

• Mass ,m = 10g
• Acceleration ,a = 5cm/s²

To Find:-

• Force acting ,F

Solution:-

Firstly we change the unit here

$:\implies$ a = 5cm/s²

$:\implies$ a = 5/100 m/s²

$:\implies$ a = 0.05m/s²

Now, converting the mass into kg

$:\implies$ m = 10/1000

$:\implies$ m = 0.01 kg

As we know that force is defined as the product of mass & acceleration .

• F = ma

where,

• F denote force
• m denote mass
• a denote acceleration

Substitute the value we get

$:\implies$ F = 0.01 × 0.05

$:\implies$ F = 0.0005 N

• Hence, the Force acting on it is 0.0005 Newton .

Answer 2

Answer:

Given :-

• A mass of 10 g is moving with an acceleration of 5 cm/s².

To Find :-

• What is the force acting on it.

Formula Used :-

$\clubsuit$ Force Formula :

$\longmapsto \sf\boxed{\bold{\pink{F =\: ma}}}$

where,

• F = Force
• m = Mass
• a = Acceleration

Solution :-

First, we have to change the acceleration from cm to m :

$\implies \sf Acceleration =\: 5\: cm/s^2$

$\implies \sf Acceleration =\: \dfrac{5}{100}\: m/s^2\: \: \bigg\lgroup \sf\bold{\green{1\: cm/s^2 =\: \dfrac{1}{100}\: m/s^2}} \bigg\rgroup$

$\implies \sf\bold{\purple{ Acceleration =\: 0.05\: m/s^2}}$

Hence, the acceleration is 0.05 m/s².

Again, we have to change the mass from g to kg :

$\implies \sf Mass =\: 10\: g$

$\implies \sf Mass =\: \dfrac{1\cancel{0}}{100\cancel{0}}\: kg\: \: \bigg\lgroup \sf\bold{\green{1\: g =\: \dfrac{1}{1000}\: kg}} \bigg\rgroup$

$\implies \sf Mass =\: \dfrac{1}{100}\: kg$

$\implies \sf\bold{\purple{Mass =\: 0.01\: kg}}$

Hence, the mass is 0.01 kg .

Now, we have to find the force acting on it :

Given :

$\bigstar$ Mass = 0.01 kg

$\bigstar$ Acceleration = 0.05 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf F =\: 0.01 \times 0.05$

$\longrightarrow \sf F =\: \dfrac{1}{100} \times \dfrac{5}{100}$

$\longrightarrow \sf F =\: \dfrac{5}{10000}$

$\longrightarrow\sf\bold{\red{F =\: 0.0005\: N}}$

$\therefore$ The force acting on it is 0.0005 N .