Answer: 2.86 m
Explanation:
To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,
ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)
In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have
mgh + 0 = 0 + KE(f)
To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have
mgh = 1/2mv² + 1/2Iw²
To get the inertia of the bodies, we use the formula
I = [m(R1² + R2²) / 2]
I = [2(0.2² + 0.1²) / 2]
I = 0.04 + 0.01
I = 0.05 kgm²
Also, the angular velocity is given by
w = v / R2
w = 4 / (1/5)
w = 20 rad/s
If we then substitute these values in the equation we have,
0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)
4.9h = 4 + 10
4.9h = 14
h = 14 / 4.9
h = 2.86 m
m = 5 kg
a = 2 m/s²
to find the force that accelerates the 4 kg object @ 2 m/s²
F = ma = 5 kg x 2 m/s² = 10 N
To find what acceleration 10 N would give a 20 kg object
a = F/m = 10 N/20 kg = 0.5 m/s
Answer:3,600 Newtons
Explanation:
The net force acting on the car is
3×10^3squared
Newtons.
Force is defined as the product of the mass of the body and its aaceleration,⇒F=ma
Substituting the above given values we get,F=(1500kg) (2.0m /s^2 squared)=3000 N=3×10^3 squared N.
N=newtons
C the runners feet pushing against the ground describes the acceleration toward the finish line
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
