Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end
of a rope is attached to the bus and the other is tied around Magnus's waist. If a force gauge placed halfway down the rope reads out a constant 2750 2750 Newtons while Magnus pulls the bus a distance of 1.60 1.60 meters, how much work does the tension force do on Magnus? The rope is perfectly horizontal during the pull.
1 answer:
Answer:
The workdone is
Explanation:
The free body diagram is shown on the first uploaded image
From the question we are given that
The force is on the force gauge
The distance that Magnus pulled the bus
Generally the workdone by the tension force on Magnus is
This negative sign show that is tension force is in the opposite direction to Magnus movement (i.e the movement of the bus )
Substituting value we have
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Answer:
F = 69.3 N
Explanation:
For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by
fr = μ N
We define a reference system parallel to the floor
block B ( lower)
Y axis
N - W₁-W₂ = 0
N = W₂ + W₂
N = (M + m) g
X axis
F -fr = M a
for block A (upper)
X axis
fr = m a (2)
so that the blocks do not slide, the acceleration in both must be the same.
Let's solve the system by adding the two equations
F = (M + m) a (3)
a =
the friction force has the formula
fr = μ N
fr = μ (M + m) g
let's calculate
fr = 0.34 (2.0 + 0.250) 9.8
fr = 7.7 N
we substitute in equation 2
fr = m a
a = fr / m
a = 7.7 / 0.250
a = 30.8 m / s²
we substitute in equation 3
F = (2.0 + 0.250) 30.8
F = 69.3 N
Answer:
The car strikes the tree with a final speed of 4.165 m/s
The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m
Explanation:
First we need to calculate the initial speed
Once we have the initial speed, we can isolate the final speed from following equation:
Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.
To do that, we replace 62 m in the first formula, as follows:
Answer:
please read the answer below
Explanation:
The angular momentum is given by
By taking into account the angles between the vectors r and v in each case we obtain:
a)
v=(2,0)
r=(0,1)
angle = 90°
b)
r=(0,-1)
angle = 90°
c)
r=(1,0)
angle = 0°
r and v are parallel
L = 0kgm/s
d)
r=(-1,0)
angle = 180°
r and v are parallel
L = 0kgm/s
e)
r=(1,1)
angle = 45°
f)
r=(-1,1)
angle = 45°
the same as e):
L = 5kgm/s
g)
r=(-1,-1)
angle = 135°
h)
r=(1,-1)
angle = 135°
the same as g):
L = 5kgm/s
hope this helps!!