A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull
parallel to the ramp’s surface. If the carton travels 5.20 m along the surface of the ramp, calculate the work done on it by (a) the rope, (b) gravity, (c) the normal force of the ramp. (d) What is the net work done on the carton? (e) Suppose that the rope is angled at 50.0∘above the horizontal, instead of being parallel to the ramp’s surface. How much work does the rope do on the carton in this case?
1 answer:
Answer:
(A) 374.4 J
(B) -332.8 J
(C) 0 J
(D) 41.6 J
(E) 351.8 J
Explanation:
weight of carton (w) = 128 N
angle of inclination (θ) = 30 degrees
force (f) = 72 N
distance (s) = 5.2 m
(A) calculate the work done by the rope
- work done = force x distance x cos θ
- since the rope is parallel to the ramp the angle between the rope and
the ramp θ will be 0
work done = 72 x 5.2 x cos 0
work done by the rope = 374.4 J
(B) calculate the work done by gravity
- the work done by gravity = weight of carton x distance x cos θ
- The weight of the carton = force exerted by the mass of the carton = m x g
- the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.
work done by gravity = 128 x 5.2 x cos 120
work done by gravity = -332.8 J
(C) find the work done by the normal force acting on the ramp
- work done by the normal force = force x distance x cos θ
- the angle between the normal force and the ramp is 90 degrees
work done by the normal force = Fn x distance x cos θ
work done by the normal force = Fn x 5.2 x cos 90
work done by the normal force = Fn x 5.2 x 0
work done by the normal force = 0 J
(D) what is the net work done ?
- The net work done is the addition of the work done by the rope, gravitational force and the normal force
net work done = 374.4 - 332.8 + 0 = 41.6 J
(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal
- work done by the rope= force x distance x cos θ
- the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp
work done = 72 x 5.2 x cos 20
work done = 351.8 J
You might be interested in
To solve this problem we will apply the concepts related to the Doppler effect. According to this concept, it is understood as the increase or decrease of the frequency of a sound wave when the source that produces it and the person who captures it move away from each other or approach each other. Mathematically this can be described as
Here,
= Original frequency
= Velocity of the observer
= Velocity of the speed
Our values are,
Using the previous equation,
Rearrange to find the velocity of the observer
Replacing we have that
Therefore the velocity of the observer is 16.2m/s
Answer:
Displacement: 6.71 m, Direction: 63.4 degrees north of east
Explanation:
In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).
Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).
And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)
And the displacement will be defined by the folliwing vector operation:
We know that the magnitude of the displacement vector is defined by the phytagoras theorem
And the angle will be defined by:
tan(beta)=3/6
beta = tan^-1(6/3)
beta = 63.43°
