Question

4) The initial velocity of a body of mass 35 gram is 15 m/s and after 10 s its velocity becomes 65cm/s. Find the magnitude of the force acting on the body.​

Answer 1

Given:

Initial velocity (u) = 15 m/s

Final velocity (v) = 65 cm/s = 0.65 m/s

Mass of body (m) = 35 g = 0.035 kg

Time interval (∆t) = 10 s

To Find:

Magnitude of force acting on the body (F)

Answer:

The rate of change of momentum of a body is directly proportional to the applied force and it takes place in the direction in which the force acts.

$\bf \implies F \propto \dfrac{\Delta p}{\Delta t} \\ \\ \rm \implies F = k \dfrac{\Delta p}{\Delta t} \\ \\ \rm k = 1 : \\ \\ \rm \implies F = \dfrac{\Delta p}{\Delta t} \\ \\ \rm \implies F = \dfrac{\Delta (mv)}{\Delta t} \\ \\ \rm \implies F = m\dfrac{\Delta v}{\Delta t} \\ \\ \rm \implies F = m\dfrac{(v - u)}{\Delta t} \\ \\ \rm \implies F = 0.035 \times \dfrac{(0.65 - 15)}{10} \\ \\ \rm \implies F = 0.035 \times (- \dfrac{14.35}{10}) \\ \\ \rm \implies F = 0.035 \times (-1.435) \\ \\ \rm \implies F = -0.05 \: N$

$\therefore$ $\boxed{\mathfrak{Magnitude \ of \ force \ acting \ on \ the \ body \ (F) = -0.05 \ N}}$

Answer 2

★Solution:-

As per the given data ,

• Mass of the body = 35 g = 0.035 kg

• Initial velocity of the body = 15 m/s

• Time taken = 10 s

• Final velocity of the body = 65 cm/s = 0.65 m/s

First we need to find the acceleration of the body in order to do that simply use first kinematic equation ,

➽ v= u + at

here ,

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

➝ a = v-u /t

➝ a = 0.65 - 15 / 10

➝ a = - 14.65 /10

➝ a =  - 1.465 m/s²

Note : here negative sign represents retardation

By using Newton's second law of motion ,

➽ F =ma

➝ F = 0.035 x - 1.46

➝ F = - 0.05 N

➝ | F | = 0.05 N

The magnitude of force acting on the body = 0.05 N