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Answer : Tension in the line = 936.7 N
Explanation :
It is given that,
Mass of student, m = 65 kg
The angle between slackline and horizontal,
The two forces that acts are :
(i) Tension
(ii) Weight
So, from the figure it is clear that :
Hence, this is the required solution.
Answer:
v₀ =3.8 m/s
Explanation:
Newton's second law of the box:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Known data
m=2.1 kg mass of the box
d= 5.4m length of the roof
θ = 20° angle θ of the roof with respect to the horizontal direction
μk= 0.51 : coefficient of kinetic friction between the box and the roof
g = 9.8 m/s² : acceleration due to gravity
Forces acting on the box
We define the x-axis in the direction parallel to the movement of the box on the roof and the y-axis in the direction perpendicular to it.
W: Weight of the box : In vertical direction
N : Normal force : perpendicular to the direction the roof
fk : Friction force: parallel to the direction to the roof
Calculated of the weight of the box
W= m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N
x-y weight components
Wx= Wsin θ= (20.58)*sin(20)° =7.039 N
Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N
Calculated of the Normal force
∑Fy = m*ay ay = 0
N-Wy= 0
N=Wy =19.34 N
Calculated of the Friction force:
fk=μk*N= 0.51* 19.34 N = 9.86 N
We apply the formula (1) to calculated acceleration of the block:
∑Fx = m*ax , ax= a : acceleration of the block
Wx-f = ( 2.1)*a
7.039 - 9.86 = ( 2.1)*a
-2.821 = ( 2.1)*a
a=(-2.821) /( 2.1)
a= -1.34 m/s²
Kinematics of the box
Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement = 5.4 m
v₀: initial speed
vf: final speed = 0
a : acceleration of the box = -1.34 m/s²
We replace data in the formula (2)
0²=v₀²+2*(-1.34)*(5.4)
2*(1.34)*(5.4)= v₀²
v₀ = 3.8 m/s
The closest answer would be C.
The choices given do not give the exact value.
To answer this, you just need to remember the main formula:
Where:
d = distance/displacement
g = acceleration due to gravity
t = time in flight
Vi = initial velocity.
With this formula, you derive all the formulas you need to look for certain components. You need to keep in mind of the following:
---if you are looking for a vertical component(y), you need to use values of vertical motion.
*Viy is always 0m/s at the beginning of a free-fall.
---If you are looking for a horizontal component(x), you need to use values of horizontal motion.
*g is always 0m/s² when taking horizontal motion into account.
--- time is the only value that is both vertical and horizontal.
Okay, let's get back to solving your problem. Let's see what your given is first:
dy = 137m (as long as it refers to height, it is vertical distance)
dx = 4km (the word, far or away usually indicates horizontal distance)
g = 1.63m/s²
The question is how fast was it going horizontally and we can derive it from our equation:
We use this because x means horizontal. But notice that we do not have time yet. So how are we going to solve this 2 variables missing? The key is that time is a horizontal and vertical component. Whatever time it took moving horizontally, it is the same vertically as well. So we use the vertical formula to derive time:
Now plug in what you know and solve for what you don't know:
The total time in flight is 12.965s.
Let's round it off to 13s.
Now that we know that, we can use this in the horizontal formula:
Hold up! Look at the unit of the horizontal distance. It is in km but all our units are expressed in m so we need to convert that first.
1km = 1,000m
4km = 4,000m
Our new horizontal distance is 4,000m.
Okay, let's wrap this up by solving for what is asked for, using all the derived values.
The horizontal velocity is 308m/s.
Such a long explanation I know, but hopefully, you learned from it.
The choices given do not give the exact value.
To answer this, you just need to remember the main formula:
Where:
d = distance/displacement
g = acceleration due to gravity
t = time in flight
Vi = initial velocity.
With this formula, you derive all the formulas you need to look for certain components. You need to keep in mind of the following:
---if you are looking for a vertical component(y), you need to use values of vertical motion.
*Viy is always 0m/s at the beginning of a free-fall.
---If you are looking for a horizontal component(x), you need to use values of horizontal motion.
*g is always 0m/s² when taking horizontal motion into account.
--- time is the only value that is both vertical and horizontal.
Okay, let's get back to solving your problem. Let's see what your given is first:
dy = 137m (as long as it refers to height, it is vertical distance)
dx = 4km (the word, far or away usually indicates horizontal distance)
g = 1.63m/s²
The question is how fast was it going horizontally and we can derive it from our equation:
We use this because x means horizontal. But notice that we do not have time yet. So how are we going to solve this 2 variables missing? The key is that time is a horizontal and vertical component. Whatever time it took moving horizontally, it is the same vertically as well. So we use the vertical formula to derive time:
Now plug in what you know and solve for what you don't know:
The total time in flight is 12.965s.
Let's round it off to 13s.
Now that we know that, we can use this in the horizontal formula:
Hold up! Look at the unit of the horizontal distance. It is in km but all our units are expressed in m so we need to convert that first.
1km = 1,000m
4km = 4,000m
Our new horizontal distance is 4,000m.
Okay, let's wrap this up by solving for what is asked for, using all the derived values.
The horizontal velocity is 308m/s.
Such a long explanation I know, but hopefully, you learned from it.