Answer:
N = 2490 N
Explanation:
Determine the upward force exerted by the horizontal platform on the box as it is lowered.
It is given that,
Mass of a box, m = 300 kg
Starting from rest at time t = 0 seconds, the box is lowered with a downward acceleration of 1.5 m/s².
It is assumed to find the upward force exerted by the horizontal platform on the box. It is given by :
mg-N=ma
So, the normal force is 2490 N.
Answer:
The maximum height reached by the body is 313.6 m
The time to return to its point of projection is 8 s.
Explanation:
Given;
initial velocity of the body, u = 78.4 m/s
at maximum height (h) the final velocity of the body (v) = 0
The following equation is applied to determine the maximum height reached by the body;
v² = u² - 2gh
0 = u² - 2gh
2gh = u²
h = u²/2g
h = (78.4²) / (2 x 9.8)
h = 313.6 m
The time to return to its point of projection is calculated as follows;
at maximum height, the final velocity becomes the initial velocity = 0
h = v + ¹/₂gt²
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
Answer:
d) v = 100.2 mph, e) t = 1.25 s, f) -0.0340
Explanation:
d) This is an exercise that we can solve using projectile launch equations, let's start by calculating the time the ball will take to take home-plate
.x = 147 t
t = x / 147
t = 60.5 / 147
t = 0.41156 s
Let's use Pythagoras' theorem to find the speed
v = √ vₓ² + ²
vₓ = dx / dt
vₓ = 147
v_{y} = dy / dt
v_{y} = 4 -16 t
We look for speed for the time of arriving at home
= 4 - 16 0.41156
v_{y} = -2,585 ft / s
v_{y} = -2.585 ft/ s ( 1 mile /5280 foot) (3600s/1h)
Let's calculate the speed
v = √ (147² + 2,585²)
v = 147.02 ft / s
v = 147.0 ft/s (1 mile/5280 feet)(3600s/1h)
v = 100.2 mph
e) the time it takes for the ball to reach the floor and = 0 foot
y = 5 + 4 t - 16 t²
0 = 5 + 4t - 16t²
t² –t / 4 -5/4 = 0
t² -0.25 t -1.25 = 0
We solve the equation and second degree
t = [0.25 ±√(0.25² + 4 1.25)] / 2
t = [0.25 ± 2.25] / 2
t₁ = 1.25 s
t₂ = -1 s
The positive time is correct
t = 1.25 s
f) The angle of speed when the ball passes home
tan θ = / vₓ
θ = tan⁺¹ (v_{y} / vx)
The distance x is given in the exercise
x = 60.5 foot
vₓ = 147 foot / s
The speed y is t = 1.25 s
v_{y} = 5 + 4 1.25 - 16 1.25²
v_{y} = -15 foot / s
θ = tan⁻¹ (-15/147)
θ = -1,947º = -0,0340 rad