Newton's third law states that whenever an object exerts a force on a second object, it exerts a force of equal magnitude and direction but in the opposite direction on the first. It is often stated as follows: Each action always opposes an equal but opposite reaction.

The subject 1 of 100kg is making a force F, to move an object from 50Kg to 2m / s ^ 2. This Force the object of 50Kg will reflect it in the opposite direction by Newton's third law.

Once the parameter of the force that both are experiencing is clarified, Newton's second law is applied to their respective calculation.

$F = ma = 50kg * 2m / s ^ 2 = 100N$

That is the force the boy exert on the man during the shove.

4 0

2 years ago

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

3 years ago