. While a person lifts a book of mass 2 kg from the floor to a tabletop, 1.5 m above the floor, how much work does the the perso
2 answers:
Answer:
29.4 J -29.4 J
Explanation:
The work done by the person on the book is given by =mgh
where m =mass
g= acceleration due to gravity
h=height
Here m=2 kg given g=9.8 kg m/ h=1.5 meter
So work done by the person =2×9.8×1.5=29.4 J
As the gravitational force work downward and person lift the book upward so work done by gravity = -mgh = -2×9.8×1.5= -29.4 J
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The work done by Joe is 0 J.
<u>Explanation</u>:
When a force is applied to an object, there will be a movement because of the applied force to a certain distance. This transfer of energy when a force is applied to an object that tends to move the object is known as work done.
The energy is transferred from one state to another and the stored energy is equal to the work done.
W = F . D
where F represents the force in newton,
D represents the distance or displacement of an object.
Force = 0 N, D = 20 cm = 0.20 m
W = 0 0.20 = 0 J.
Hence the work done by Joe is 0 J.
Answer:
Diagrams in pictures
Explanation:
Using energy I can get
m g h = 1/2 m v^2
So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.
(H= 1,3m sin25)
V=2,32m/s
V= a t
And
X= v t +1/2 a t^2
Knowing v=2,32 m/s and x= 1,3 m
I can get
a= 6,21m/s2
F= m a
I can get the force of the box when it collides with the spring
F= 12, 425 N
The force the spring makes on the box then is
F = -12,425N = -k d
Then the spring's constant is k= 51,75N/m
To make the two diagrams I need the functions of time when the box slows down
I use the same two equations
V= a t
And
X= v t + 1/2 a t^2
Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.
I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).
Then,
V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec
X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.
for time between 0 and 0,689 sec
Diagrams and equations are in the pictures
Answer:
(a) θ = 33.86°
(b) Ay = 49.92 N
Explanation:
You have that the magnitude of a vector is A = 89.6 N
The x component of such a vector is Ax = 74.4 N
(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:
(1)
Ax: x component of vector A
A: magnitude of vector A
θ: angle between vector A and the x axis
You solve the equation (1) for θ, by using the inverse of cosine function:
the angle between the A vector and the x axis is 33.86°
(b) The y component of the vector is given by:
the y comonent of the vecor is Ay = 49.92 N