Question:

A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25 sec. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Solution:

Calculating the Acceleration:

Initial velocity of the body $${u=2 \ \frac{m}{s}}$$

Final velocity of the body $${v=3.5 \ \frac {m}{s}}$$

Time taken t=25 sec

Acceleration $${a=\frac{v-u}{t}}$$

$${a=\frac{3.5-2}{25}}$$

$${a=\frac{1.5}{25}}$$

$${a=0.06 \ \frac{m}{s^2}}$$

Calculating the force:

As per Newton’s second law $${F=ma}$$

Mass of the body m= 3 Kg

Acceleration a=0.06 ms^-2

$${F=ma =(3 \ kg)(0.06 \ \frac{m}{s^2})}$$

$${F=0.18 \ N}$$

Direction of the force:

The direction of the force is same as that of the direction of the body’s motion.

Physics Quiz:

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