Question

a motor car of 1000kg is moving with uniform velocity of 5m/s its final velocity becomes 25m/s in 4s due to applied force calculate the force

Answer 1

$\bigstar\large\boxed{\mathtt{\green{Given:}}}$

• Mass of the motor car = 1000 kg
• Initial velocity = 5 m/s
• Final velocity = 25 m/s
• Time taken = 4s

$\bigstar\large\boxed{\mathtt{\blue{To \: find :}}}$

• Force applied

$\bigstar\large\boxed{\mathtt{\pink {How \: to \: solve :}}}$

• In order to find the force applied on the car we need to know the acceleration of the car. A s the values of v, u, and t has already been given to us we can simply substitute these values in the first equation of motion in order to get the answer

$\bigstar\large\boxed{\mathtt{\orange {Solution :}}}$

By using the first equation of motion ,

$\implies\boxed{\mathtt{v = u+ at } }$

here ,

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

Substituting the given values in the above equation ,

$\implies\mathtt{25= 5+ a\times4 }$

$\implies\mathtt{25- 5= a\times4 }$

$\implies\mathtt{20= a\times4 }$

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

$\implies\mathtt{a = 5 \dfrac{m }{s^{2}} }$

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We know that ,

$\implies\boxed{ \mathtt{ F = ma } }$

here ,

• F = force applied
• m = mass of the motor car
• a = acceleration

$\implies\mathtt{ F = 1000 \times 5 }$

$\implies\mathtt{\red { F = 5000 N }}$

The force applied on the motor car is 5000 N

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$\bigstar\large\boxed{\mathtt{\orange {Additional \: formula:}}}$

Second equation of motion ,

$\implies\mathtt{ s = ut + \dfrac{1}{2}at ^{2} }$

Third equation of motion ,

$\implies\mathtt{v^{2}=u^{2} + 2as }$

Distance covered by the object in n th second,

$\implies\mathtt{ s_n = u+ \dfrac{a}{2}(2n-1) }$