Answer:
v_f = 1.05 m/s
Explanation:
From conservation of energy;
E_f = E_i
Thus,
(1/2)m(v_f)² + (1/2)I(ω_f)² + m•g•h_f + (1/2)k•(x_f)² = (1/2)m(v_i)² + (1/2)I(ω_i)² + m•g•h_i + (1/2)k•(x_i)²
This reduces to;
(1/2)m(v_f)² + (1/2)Ik(x_f)² = (1/2)k•(x_i)²
Making v_f the subject, we have;
v_f = [√(k/m)] * [√((x_i)² - (x_f)²)]
We know that ω = √(k/m)
Thus,
v_f = ω[√((x_i)² - (x_f)²)]
Plugging in the relevant values to obtain;
v_f = 17.8[√((0.068)² - (0.034)²)]
v_f = 17.8[0.059] = 1.05 m/s
In a), no work is done to move horizontally 'across" gravity. In b), he does (m) (g) (h) = (100)(9.8)(1) = 980 joules of work to lift the bag, and then no more work is done by the belt to move both of them horizontally.
Answer:
Option E is correct.
There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.
Explanation:
Normally, ignoring air resistance, for projectile motion, the range (horizontal distance teavelled) of the motion is given as
R = (u² sin 2θ)/g
where
u = initial velocity of the projectile = 20 m/s
θ = angle above the horizontal at which the projectile was launched = 30°
g = acceleration due to gravity = 9.8 m/s²
R = (30² sin 60°) ÷ 9.8
R = 78.53 m
So, Normally, the stone should travel a horizontal distance of 78.53 m. So, travelling a horizontal distance of 32 m (less than half of what the range should be without air resistance) means that, the motion of the stone was impeded, hence, option E is correct.
There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.
Hope this Helps!!!
