Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.
Answer:
<em>The distance is now 4d</em>
Explanation:
<u>Mechanical Force</u>
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
F = m.a
Where a is the acceleration of the object.
The acceleration can be calculated by solving for a:
Once we know the acceleration, we can calculate the distance traveled by the block as follows:
If the block starts from rest, vo=0:
Substituting the value of the acceleration:
Simplifying:
When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:
And the distance is now:
Dividing d'/d:
Simplifying:
Thus:
d' = 4d
The distance is now 4d
We have 
. So, 
.
. So 
.
Thus we can convert the units of the given quantity.
That is,
.
The quantity is converted to the required units.
<span>Her center of mass will rise 3.7 meters.
First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so
8.5 m / 9.8 m/s^2 = 0.867346939 s
And the distance a object under constant acceleration travels is
d = 0.5 A T^2
Substituting known values, gives
d = 0.5 9.8 m/s^2 (0.867346939 s)^2
d = 4.9 m/s^2 * 0.752290712 s^2
d = 3.68622449 m
Rounded to 2 significant figures gives 3.7 meters.
Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.</span>
12.00 min = 0.2 hr
8.00 min = 0.15 hr
Total distance:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)
= 8.25 km
Average speed:
(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3
= 15 km/hr
Change in position:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)
= 0.25 km
Average velocity:
(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3
≈ 1.67 m/s
