Question

Force of 20n acts of on the body of the mass 4kg at rest force 10second calculated the velocity aquared and distance travelled by the body during the intervel

Answer 1

Appropriate Question:

A force of 20N acts of on the body of the mass 4kg at rest for 10 seconds. Calculate the velocity acquired and distance travelled by the body during the interval.

Explanation:

As per the provided information in the given question, we have been provided that the mass of the body is 4kg (which is at rest) on which a force of 20N acts for 10 seconds. And, we've been asked to calculate velocity acquired i.e, final velocity and distance travelled.

As per the statements, we have :

• Mass of the body, m = 4 kg
• Initial velocity, u = 0 m/s [At rest]
• Time taken, t = 10 seconds
• Force applied, f = 20 N

To Find : Final velocity [v] & Distance travelled [s].

»› F I N A L ⠀V E L O C I T Y :

By using the formula to calculate force that is,

$\qquad \qquad \Bigg [ \textbf{\textsf{F = ma}}\Bigg ]$

• F denotes Force
• m denotes mass
• a denotes acceleration

Note : Acceleration is the rate for change in velocity. And mathematically, a = (v ― u)/t.

$\twoheadrightarrow\sf{20=4a}\\ \\\twoheadrightarrow\sf{20=4\Bigg \lgroup \dfrac{v-u}{t} \Bigg \rgroup }\\ \\\twoheadrightarrow\sf{20=4\Bigg \lgroup \dfrac{v-0}{10} \Bigg \rgroup }\\ \\ \twoheadrightarrow\sf{\cancel{\dfrac{20}{4}}=\Bigg \lgroup \dfrac{v-0}{10} \Bigg \rgroup } \\ \\\twoheadrightarrow\sf{5= \dfrac{v}{10} }\\ \\ \twoheadrightarrow\sf{5 \times 10= v }\\ \\ \twoheadrightarrow \underline{\boxed{\pmb{\frak{50 \; ms^{-1}= v }}}}$

∴ The final velocity of the body is 50 m/s.

$\rule{200}2$

»› D I S T A N C E ⠀ T R A V E L L E D :

By using the third equation of motion,

$\qquad \qquad \Bigg [ \textbf{\textsf{v}}^\textbf{\textsf{2}} - \textbf{\textsf{u}}^{\textbf{\textsf{2}}} = \textbf{\textsf{2as}}\Bigg ]$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

Acceleration :

$\twoheadrightarrow\sf{ \Bigg [ \textsf{\textbf{v=u+at}} \Bigg ]}\\\\ \twoheadrightarrow\sf{\dfrac{v-u}{t} = a}\\ \\\twoheadrightarrow\sf{\dfrac{50-0}{10} = a}\\\\\twoheadrightarrow\sf{\dfrac{50}{10} = a}\\ \\\twoheadrightarrow \underline{\boxed{\frak{ 5\; ms^{-2}= a }}}$

Now, substitute the values in their equation of motion.

$\twoheadrightarrow\sf{(50)^2 - (0)^2 = 2(5)(s)}\\ \\ \twoheadrightarrow\sf{2500 - 0 = 10s}\\ \\\twoheadrightarrow\sf{2500= 10s}\\ \\\twoheadrightarrow\sf{\cancel{\dfrac{2500}{10} }= s}\\ \\ \twoheadrightarrow \underline{\boxed{\pmb{\frak{250 \; m= s }}}}$

∴ The distance travelled by the body is 250 m.

$\rule{200}2$

Answer 2

Answer:

Given :-

• A force of 20 N acts of the body of the mass 4 kg at rest force 10 seconds.

To Find :-

• What is velocity acquired and distance travelled by the body during the interval.

Formula Used :-

$\clubsuit$ Force Formula :

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

where,

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

$\clubsuit$ First Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{v =\: u + at}}}\: \: \: \bigstar$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

$\clubsuit$ Second Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2} at^2}}}\: \: \: \bigstar$

where,

• s = Distance Travelled
• u = Initial Velocity
• t = Time Taken
• a = Acceleration

Solution :-

First, we have to find the force :

Given :

• Force = 20 N
• Mass = 4 kg

According to the question by using the formula we get,

$\implies \bf F =\: ma$

$\implies \sf 20 =\: 4 \times a$

$\implies \sf 20 =\: 4a$

$\implies \dfrac{20}{4} =\: a$

$\implies \sf 5 =\: a$

$\implies \sf\bold{\blue{a =\: 5\: m/s^2}}$

Hence, the acceleration is 5 m/s² .

Now, we have to find the velocity acquired :

Given :

• Initial Velocity = 0 m/s
• Acceleration = 5 m/s²
• Time = 10 seconds

According to the question by using the formula we get,

$\mapsto \bf v =\: u + at$

$\mapsto \sf v =\: 0 + (5)(10)$

$\mapsto \sf v =\: 0 + 5 \times 10$

$\mapsto \sf v =\: 0 + 50$

$\mapsto \sf\bold{\purple{v =\: 50\: m/s}}$

Hence, the final velocity acquired is 50 m/s .

Now, we have to find the distance travelled by the body during the interval :

Given :

• Time = 10 seconds
• Acceleration = 5 m/s²
• Initial Velocity = 0 m/s

According to the question by using the formula we get,

$\leadsto \bf s =\: ut + \dfrac{1}{2} at^2$

$\leadsto \sf s =\: (0)(10) + \dfrac{1}{2} \times (5)(10)^2$

$\leadsto \sf s =\: 0 \times 10 + \dfrac{1}{2} \times 5 \times (10 \times 10)$

$\leadsto \sf s =\: 0 + \dfrac{1}{\cancel{2}} \times {\cancel{500}}$

$\leadsto \sf s =\: 0 + 250$

$\leadsto \sf\bold{\red{s =\: 250\: m}}$

Hence, the distance travelled by the body during the interval is 250 m .

$\therefore$ The final velocity acquired is 50 m/s and the distance travelled by the body during the interval is 250 m .