A box of mass 60 kg has two horizontal forces acting on it. The first force pushes from the left and has magnitude F1 = 130 N. The other force drags the box from the right and has magnitude F2. The ground is rough and horizontal, and the box speeds up with an acceleration of 3.7 m/s? in the direction of the applied forces. If the magnitude of the force of friction is 50 N, what is the value of F2? (Arrows representing forces are not drawn to scale).
A box of mass 60 kg has two horizontal forces acting on it. The first force pushes from the left and has magnitude F1 = 130 N. The other force drags the box from the right and has magnitude F2. The ground is rough and horizontal, and the box speeds up with an acceleration of 3.7 m/s? in the direction of the applied forces. If the magnitude of the force of friction is 50 N, what is the value of F2? (Arrows representing forces are not drawn to scale).
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Question 16, Physics - equation sheet attached

Transcribed Image Text:Physics 114 Equation Sheet
Constants and Conversions
Kinematics Continued
g = 9.80 m/s
Free-fall acceleration
Δν
Instantaneous
ainst. = lim
At-o At
Acceleration
1N = 1 kg m/s?
Newton
Uniform motion
(v) = (v); = constant
Position in uniform
X = x + (v)At
Mathematics, Scaling and Vectors
b = a* + loga (b) = x
motion
Logarithm
Constant
(v); = (v,); + azAt
acceleration:
1
log(ab) = log(a) + log (b)
x, = x, + (v,),At +a, (at)?
log Ax" = n log x + log A
(v,); = (v,)} + 2a,Ax
Volume of a sphere
V =
Surface area of a
A = 4ar?
Forces
sphere
Newton's second law
Fnet = EF = mã
%3D
Volume of a cylinder
V = arl
Newton's second law
Fnetx = EF = ma,
%3D
Surface area of a
A = 2ar? + 2rl
Component form
Fnety = ER, = may
cylinder
Mass density
p = m/V
Newton's Third Law
FA en =-
ton A
A, = A cos e (rel. to x-axis)
Weight
w = mg
x -component of a
vector Å
Apparent weight
Wapp = magnitude of supporting forces
y -component of a
Ay = A sin 8 (rel to x-axis)
vector Å
Kinetic friction
fk = Han
Magnitude of vector Ả
Static friction
A = JA + A,
Reynolds number
Re = pvl/n
Direction of A relative
8 = tan-(Ay/A,)
1
Drag (high Reynolds
number)
=CopAv?
to x-axis
Addition of two vectors If = Å + B, then
C, = A, + B,
D = 6nyrv
Drag (low Reynolds
number)
Cy = Ay + By
Circular Motion
Kinematics
Centripetal acceleration
a =
Displacement
Ax = x - X
Average Velocity
Ax
Frequency
1
Varg
T
2ar
At
Ax
Vinst. = lim
Instantaneous Velocity
At+0 At
Av
Average Acceleration
davg
Δε

Transcribed Image Text:A box of mass 60 kg has two horizontal forces acting on it. The first force pushes from the left and
has magnitude F1 = 130 N. The other force drags the box from the right and has magnitude F2. The
ground is rough and horizontal, and the box speeds up with an acceleration of 3.7 m/s? in the
direction of the applied forces.
If the magnitude of the force of friction is 50 N, what is the value of F2? (Arrows representing forces
are not drawn to scale).
O 170 N
O 220 N
O 80 N
O 140 N
O 90 N
Expert Solution

Step 1
mass = m = 60 kg
acceleration = a = 3.7 m/s2
F1=130 N
frictional force = f = 50 N
F2= ?
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