How much force does it take to accelerate a 50.8 kg person at 3.50 m/s^2?
1 answer:
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The spring constant is 181.0 N/m
Explanation:
We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:
where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where
k is the spring constant of the spring
x = 2.08 cm = 0.0208 m is the compression of the spring
m = 12.3 g = 0.00123 kg is the mass of the dart
is the acceleration due to gravity
h = 3.25 m is the maximum height of the dart
Solving for k, we find:
Learn more about potential energy:
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Answer:
30.67m
Explanation:
Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;
s = ut ± ------------(i)
Where;
s = horizontal/vertical displacement
u = initial horizontal/vertical component of the velocity
a = acceleration of the projectile
t = time taken for the projectile to reach a certain horizontal or vertical position.
Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;
h = vt ± ------------(ii)
Where;
h = vertical displacement
v = initial vertical component of the velocity = usinθ
a = acceleration due to gravity (since vertical motion is considered)
t = time taken for the projectile to hit the target
<em>From the question;</em>
u = 47m/s, θ = 0.6rads
=> usinθ = 47 sin 0.6
=> usinθ = 47 x 0.5646 = 26.54m/s
t = 1.7s
Take a = -g = -10.0m/s (since motion is upwards against gravity)
Substitute these values into equation (ii) as follows;
h = vt -
h = 26.54(1.7) -
h = 45.118 - 14.45
h = 30.67m
Therefore, the vertical distance is 30.67m