6300
N
Explanation:
We will employ the following equation of kinematics
x
=
1
2
a
t
2
which describes an object traveling in one dimension with a constant acceleration and an initial velocity of zero.
We know the displacement and the time elapsed so we may solve for the acceleration:
40.0
m
=
1
2
a
(
3.0
s
)
2
Solving for
a
yields:
a
=
8.89
m
s
2
Now, knowing the acceleration of the car as well as the mass of the car we can apply Newton's second law, which states
F
net
=
m
a
All we need to do is plug in our values for
m
and
a
:
F
net
=
710
kg
⋅
8.89
m
s
2
=
6311.9
N
Considering the fact that our final answer should have only two significant digits, we will round to the nearest hundred:
F
net
Answer:
A) 3.48m/s
B) 3.92m
C) 2.32m
D 23.33m/s
Explanation:
ma(t)=mg-At
a(t)= g - (At/m)
V(t)= integrala(T)dT = gt- (At^2/2m)
Initial x coordinator of the box is zero
X(t)= integralV(t)dt= 1/2gt^2-(At^3/6m)
a) V =( 9.81×1) -(38×1^2/2×3)
V= 9.81-6.33= 3.48m/s
b)-AT^2/2m + gT= 0
T=2mg/A= (2×3×9.81)/38
T= 1.549m
X(T)= (1/2×9.81×1.549^3)- (38×1.549^3/6×3)
X(T)= 11.768- (141.23/18) = 11.768 - 7.85= 3.92m
C) 1/2gT''^2 - AT''^3/6m =0
The only non trivial solution is T''= 3mg/A
T=(3×3×9.81)/38 = 2.32m
D) V = 9.81×3) - (38×3^2/6)
V= 29 - 5.667= 23.33m/s
