Question

A force acting on a particle of mass 350g displaces it through 500cm in 2s. Find the magnitude of the force if the initial velocity of the particle is zero.

Answer 1

Answer:

answer is 0.70 N

Explanation:

m= 350g= 0.35kg

s= 500cm = 5cm

t= 2s

u= 0m/s

find f...

by second law of motion:

s=ut+1/2at^2

=0+1/2×a×4

a=2m/s^2

by using equation of motion:

F=ma

=0.35×2

=0.70 N

Answer 2

Answer :

➥ The magnitude of the force = 0.875 N

Given :

➤ Mass of the partical (m) = 350 g

➤ Displacement of the partical (s) = 500 cm

➤ Time taken by the partical (t) = 2 sec

➤ Intial velocity of the partical (u) = 0 m/s

To Find :

➤ Magnitude of the force (f) = ?

Solution :

◈ Mass (m) = 350 g = 0.35 kg

◈ Displacement (s) = 500 cm = 5 m

First we need to find acceleration then we find magnitude of the force.

From second equation of motion

$\tt{: \implies s = ut + \dfrac{1}{2}a {t}^{2} }$

$\tt{: \implies 5 = 0 \times 2 + \dfrac{1}{ \cancel{2}} \times a \times \cancel{2 }\times 2 }$

$\tt{: \implies 5 = 0 + 1 \times a \times 2}$

$\tt{: \implies 5 = 0 + a \times 2}$

$\tt{: \implies 5 = 0 + 2a}$

$\tt{: \implies 5 = 2a}$

$\tt{: \implies \dfrac{5}{2} = a}$

$\tt{: \implies \purple{ \underline{ \overline{ \boxed{ \green{ \bf{ \: \: 2.5 \: m/s^2 \: \: }}}}}}}$

Now ,

We find magnitude of force

From second low of Newton

$\tt{: \implies F = ma}$

$\tt{: \implies F = 0.35 \times 2.5}$

$\tt{: \implies \green{ \underline{ \overline{ \boxed{ \purple{ \bf{ \: \: F = 0.875 \: N\: \: }}}}}}}$

Hence, the magnitude of the force is 0.875 N.