Question

Two trucks of mass equal to 5000kg are moving on a road. One is moving with velocity of 50m/s while the other is moving with a velocity of 2m/s. Calculate the force required to stop them in 20seconds? Which one of them requires more force and why?

Answer 1

Answer:

Two trucks of mass equal to 5000kg are moving on a road. One is moving with velocity of 50m/s while the other is moving with a velocity of 2m/s. Calculate the force required to stop them in 20seconds? Which one of them requires

Answer 2

Solution:

Truck 1 :

as per the given data ,

• mass = 5000 kg
• initial velocity = 50 m/s
• final velocity = 0 m/s (as it comes to rest )
• time taken by the truck to come to rest = 20 s

First we need to find the acceleration of the truck in order to do this lets use the first equation of motion

$\bigstar\boxed{\mathtt{\pink{v=u+at}}}$

here ,

• v= final velocity
• u = initial velocity
• a = acceleration
• t= time taken

Now substituting the given values in the above equation we get ,

$\rightarrow\mathtt{0=50+a \times 20}$

$\rightarrow\mathtt{a=\dfrac{-50}{20}}$

$\rightarrow\mathtt{\red{a= - 2.5\dfrac{m}{s^{2} }}}$

Note : here the negative sign denotes retardation

_____________________

we know that ,

$\bigstar\boxed{\mathtt{\pink{F=ma}}}$

here ,

• F= force
• m= mass
• a = acceleration

$\rightarrow\mathtt{F=5000 \times 2.5}$

$\rightarrow\mathtt{\red{F=12,500\: N}}$

Truck 2 :

as per the given data ,

• mass = 5000 kg
• initial velocity = 2 m/s
• final velocity = 0 m/s (as it comes to rest )
• time taken by the truck to come to rest = 20 s

First we need to find the acceleration of the truck in order to do this lets use the first equation of motion

$\bigstar\boxed{\mathtt{\pink{v=u+at}}}$

here ,

• v= final velocity
• u = initial velocity
• a = acceleration
• t= time taken

Now substituting the given values in the above equation we get ,

$\rightarrow\mathtt{0=2+a \times 20}$

$\rightarrow\mathtt{a=\dfrac{-2}{20}}$

$\rightarrow\mathtt{\red{a= - 0.1\dfrac{m}{s^{2} }}}$

Note : here the negative sign denotes retardation

_____________________

we know that ,

$\bigstar\boxed{\mathtt{\pink{F=ma}}}$

here ,

• F= force
• m= mass
• a = acceleration

$\rightarrow\mathtt{F=5000 \times 0.1 }$

$\rightarrow\mathtt{\red{F=500\: N}}$

Answer:

• The force required to stop truck 1 is 12,500 N

• The force required to stop truck 2 is 500 N

hence from this we can conclude that the force required to stio truck 1 is greater than truck 2