Question

Find the force required to move a body weighting 200kg on a rough horizontal surface having cofficient of friction 0.35​

Answer 1

Answer:

The force required to move the body is 686 N.

Step-by-step explanation:

Mass of the body = 200 kg

Coefficient of friction = 0.35

Force required to move the body = ??

$\bigstar \: \boxed{\sf{F = \mu N}}$

$\sf{N = {m \times g}}$ so, replacing it in the formula, it forms,

$\sf{F =\mu \times \: m \times g}$

• $\mu$ = 0.35
• m = 200 kg
• g = 9.8 m/s²

$\longrightarrow{\sf{F = 0.35 \times 200 \times 9.8}}$

$\longrightarrow{\sf{F = 70 \times 9.8}}$

$\longrightarrow{\sf{F = 686 \: N}}$

Force required = 686 N

Therefore, the force required to move the body is 686 N.

Answer 2

Answer:

Given :-

• A body weighing 200 kg on a rough horizontal surface having coffecient of friction of 0.35.

To Find :-

• What is the force required to move a body.

Formula Used :-

$\clubsuit$ Force Of Friction Formula :

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

where,

• $\mu$ = Friction
• m = Mass
• g = Acceleration due to gravity

Solution :-

Given :

$\bigstar \: \: \: \sf Friction (\mu) =\: 0.35$

$\bigstar \: \: \sf Mass (m) =\: 200\: kg$

$\bigstar \: \: \sf Acceleration\: due\: to\: gravity (g) =\: 9.8\: m/s^2$

According to the question by using the formula we get,

$\longrightarrow \sf F =\: 0.35 \times 200 \times 9.8$

$\longrightarrow \sf F =\: \dfrac{35}{100} \times 20{\cancel{0}} \times \dfrac{98}{1\cancel{0}}$

$\longrightarrow \sf F =\: \dfrac{35}{100} \times 20 \times 98$

$\longrightarrow \sf F =\: \dfrac{35}{100} \times 1960$

$\longrightarrow \sf F =\: \dfrac{686\cancel{00}}{1\cancel{00}}$

$\longrightarrow \sf F =\: \dfrac{686}{1}$

$\longrightarrow \sf\bold{\red{F =\: 686\: N}}$

$\therefore$ The force required to move a body is 686 N .