The net force on the system:
F = m₂g - m₁gsin(∅)
F = 39.5 x 9.81 - 43 x 9.81 x sin(30)
F = 176.58 N
Now, we use F = ma to find the acceleration on each mass.
F = m₁a₁
a₁ = 176.58 / 43
a₁ = 4.11 m/s²
F = m₂a₂
a₂ = 176.58 / 39.5
a₂ = 4.47 m/s²
Answer:
acceleraions 5.76g and 20.55g
Explanation:
This constant acceleration exercise can be solved using the kinematic equations in one dimension
Vf = Vo + a t
As part of the rest Vo = 0
a = Vf / t
a = 282/5
a = 56.4 m / s2
In relation to the acceleration of gravity
a ’= a / g = 56.4 / 9.8
a ’= 5.76g
To calculate the acceleration to stop we use the same formula
a2 = 282 / 1.40
a2 = 201.4 m / s2
This acceleration of gravity acceleration function is
a2 ’= 201.4 / 9.8
a2 ’= 20.55g
Walking at a speed of 2.1 m/s, in the first 2 s John would have walked
(2.1 m/s) (2 s) = 4.2 m
Take this point in time to be the starting point. Then John's distance from the starting line at time <em>t</em> after the first 2 s is
<em>J(t)</em> = 4.2 m + (2.1 m/s) <em>t</em>
while Ryan's position is
<em>R(t)</em> = 100 m - (1.8 m/s) <em>t</em>
where Ryan's velocity is negative because he is moving in the opposite direction.
(b) Solve for the time when they meet. This happens when <em>J(t)</em> = <em>R(t)</em> :
4.2 m + (2.1 m/s) <em>t</em> = 100 m - (1.8 m/s) <em>t</em>
(2.1 m/s) <em>t</em> + (1.8 m/s) <em>t</em> = 100 m - 4.2 m
(3.9 m/s) <em>t</em> = 95.8 m
<em>t</em> = (95.8 m) / (3.9 m/s) ≈ 24.6 s
(a) Evaluate either <em>J(t)</em> or <em>R(t)</em> at the time from part (b).
<em>J</em> (24.6 s) = 4.2 m + (2.1 m/s) (24.6 s) ≈ 55.8 m
Answer:
Yes
Explanation:
The spring force is given as:
F = kd
F is the spring force
K is the spring constant
d is the magnitude of the stretch
Since k is a constant, therefore, doubling the stretch distance will double the force.
Both stretch distance and force applied can be said to be directly proportional to one another.
