Question

A meter stick whose mass is 290 grams lies on ice. You pull at one end of the meter stick, at right angles to the stick, with a force F= 9 newtons. The ensuing motion of the meter stick is quite complicated, but what are the initial magnitude and direction of the rate of change of the momentum of the stick, dPtot/d⁢t, when you first apply the force?
Magnitude N
Direction leftdownrightup
What is the magnitude of the initial acceleration of the center of the stick?
m/s2

Answer 1

Answer

given,

mass of the stick = 290 grams = 0.29 Kg

Force on the stick on one side = F =  9 N

force acting perpendicular to stick.

magnitude of acceleration

rate of change of angular momentum is equal to Force

rate of change of angular momentum = 9 N

F = m a

$a = \dfrac{F}{m}$

$a = \dfrac{9}{0.29}$

a = 31.034 m/s²

Direction of motion will in the direction of force application or in the direction of change of velocity

Answer 2

The magnitude of the initial acceleration is mathematically given as

a = 31.034 m/s^2

What is the magnitude of the initial acceleration of the center of the stick?

Question Parameters:

A meter stick whose mass is 290 grams lies on ice

at right angles to the stick, with a force F= 9 newtons.

Generally, the equation for the Force   is mathematically given as

F = m a

a=F/m

Therefore

a=9/0.19

a = 31.034 m/s^2

In conclusion, the initial acceleration is

a = 31.034 m/s^2

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