a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.
By Newton's second law, at the moment right before the system starts to move,
• net horizontal force
∑ F[h] = F[push] - F[s. friction] = 0
• net vertical force
∑ F[v] = F[normal] - F[weight] = 0
and we have
F[s. friction] = µ[s] F[normal]
It follows that
F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N
where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of
(1068.66 N) / g ≈ 110 kg
b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to
∑ F[h] = F[push] - F[k. friction] = 0
∑ F[v] = F[normal] - F[weight] = 0
so that
F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N
and so the minimum force needed to keep the system moving is
F[push] = 710.45 N ≈ 710 N
Answer:
Answer:This organism may be identified by its color, the spines on its back, the antennae, and therefore the long, thin body. There are many other characteristics that might even be wont to identify this organism.
Explanation:
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
Frequency = (speed) / (wavelength) = 20 / 0.5 = 40 per second = 40 Hz.
