A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of
0.5 m/s2 across a horizontal floor. What is the value of the normal force on the box? 200 N 243 N 156 N 225 N
1 answer:
Answer:
225 N
Explanation:
"Below the horizontal" means he's pushing down at an angle.
Draw a free body diagram of the box. There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.
Sum of forces in the y direction:
∑F = ma
N − mg − F sin θ = 0
N = F sin θ + mg
Plug in values:
N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)
N = 225 N
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Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.