The net force on the block perpendicular to the floor is
∑ F[perp] = F[normal] - mg = 0
so that
F[normal] = (5 kg) g = 49 N
Then
F[friction] = 0.1 F[normal] = 4.9 N
so that the net force parallel to the floor is
∑ F[para] = -4.9 N = (5 kg) a
Solve for the acceleration a :
a = (-4.9 N) / (5 kg) = -0.98 m/s²
Starting with an initial velocity of 5 m/s, the box comes to a stop after time t such that
0 = 5 m/s - (0.98 m/s²) t
⇒ t ≈ 5.1 s
This is another time to look at Newton's 2nd law of motion:
Net Force = (mass) x (acceleration)
If the object is not moving, then its acceleration is certainly zero, and Newton's law looks like this:
Net Force = (mass) x (zero)
or Net Force = (zero) .
"Net Force = zero" means that if there ARE any forces acting on the object, then they add up to zero, and we call them "balanced" forces.
So the answer is '<em>yes</em>', and that's why.
La velocidad vertical del tanque después de caer 10 m es 14 m/seg .
La velocidad vertical del tanque se calcula mediante la aplicación de la fórmula de velocidad , la componente vertical Vfy, del movimiento horizontal como se muestra a continuación :
Vfy=?
h = 10 m
Fórmula de Velocidad vertical Vfy:
Vfy² = 2*g*h
Vfy= √(2*9.8m/seg2* 10m )
Vfy= 14 m/seg
Answer:
14.2 m
Explanation:
Using conservation of energy:
PE at top = KE at bottom
mgh = ½ mv²
h = v² / (2g)
h = (16.7 m/s)² / (2 × 9.8 m/s²)
h = 14.2 m
Using kinematics:
Given:
v₀ = 16.7 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 14.2 m
