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
The car has slowed down because of the deceleration or decelerating force acting on it. The decelerating force acting on it is generated by the application of brakes or the frictional force acting in the opposite direction of car's motion.
Answer:
-20.0 m/s and 30.0 m/s
Explanation:
Momentum is conserved:
m (30.0) + m (-20.0) = m v₁ + m v₂
30.0 − 20.0 = v₁ + v₂
10.0 = v₁ + v₂
Since the collision is perfectly elastic, energy is also conserved. Since there's no rotational energy or work done by friction, the initial kinetic energy equals the final kinetic energy.
½ m (30.0)² + ½ m (-20.0)² = ½ mv₁² + ½ mv₂²
(30.0)² + (-20.0)² = v₁² + v₂²
1300 = v₁² + v₂²
We now have two equations and two variables. Solve the system of equations using substitution:
1300 = v₁² + (10 − v₁)²
1300 = v₁² + 100 − 20v₁ + v₁²
0 = 2v₁² − 20v₁ − 1200
0 = v₁² − 10v₁ − 600
0 = (v₁ + 20) (v₁ − 30)
v₁ = -20, 30
If v₁ = -20, v₂ = 30.
If v₁ = 30, v₂ = -20.
So either way, the final velocities are -20.0 m/s and 30.0 m/s.
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
