Two forces act on a 41-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other ha
s a magnitude 35 N at 32° clockwise from the positive y-axis. What is the magnitude of this object's acceleration?
1 answer:
Answer:
a = 1.41 m/s²
Explanation:
Given that
mass ,m= 41 kg
F₁ = 65 N , θ = 59°
F₂ = 35 N ,θ = 32°
The component of Force F₁
F₁x= F₁cos59° i
F₁x= 65 x cos59° i = 33.47 i
F₁y= - F₁ sin 59° j
F₁y= - 65 x sin 59° j = - 55.71 j
The component of Force F₂
F₂x= F₂ sin 32° i
F₂x= 35 x sin 32° i = 18.54 i
F₂y= F₂ cos 32° j
F₂y= 35 x cos 32° j = 29.68 j
The total force F
F= 33.47 i + 18.54 i - 55.71 j + 29.68 j
F= 52.01 i - 26.03 j
The magnitude of the force F
F=58.16 N
We know that
F= m a
a= Acceleration
m=mass
58.16 = 41 x a
a = 1.41 m/s²
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a. It starts at point B.
vp = 2.53*10⁴ m/s
a. it starts at point A.
ve= 1.08*10⁶ m/s
Explanation:
a) As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.
Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:
ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)
⇒
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.
Replacing by these values, and solving for v, we have:
⇒ vp = 2.53*10⁴ m/s
b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:
First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.
Second, as its charge is (-e) the change in electric potential energy had been negative also:
ΔUe = -e*ΔV = -e* (VB-VA)
In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:
where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.
Replacing by these values, and solving for v, we have:
⇒ ve = 1.08*10⁶ m/s