4. What relationship between gravitational
2 answers:
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a) See free-body diagram in attachment
b) The book is stationary in the vertical direction
c) The net horizontal force is 35 N in the forward direction
d) The net force on the book is 35 N in the forward horizontal direction
e) The acceleration is in the forward direction
Explanation:
a)
The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.
From the diagram of this book, we see there are 4 forces acting on the book:
- The applied force, F = 50 N, pushing forward in the horizontal direction
- The frictional force, , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)
- The weight of the book, , where m is the mass of the book and
is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:
- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book
b)
The book is in equilibrium in the vertical direction, therefore there is no motion.
In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is
where a is the acceleration; however, since N = W, this becomes
And since the book is initially at rest on the desk, this means that there is no motion.
c)
We said there are two forces acting in the horizontal direction:
- The applied force, F = 50 N, forward
- The frictional force, , backward
Since they act along the same line, we can calculate their resultant as
and therefore the net force is 35 N in the forward direction.
d)
The net force is obtained as the resultant of the net forces in the horizontal and vertical direction. However, we have:
- The net force in the horizontal direction is 35 N
- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction
Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.
e)
The acceleration of the book can be calculated by using Newton's second law:
where
is the net force
m is the mass
a is the acceleration
Here we have:
(in the forward direction)
m = 4 kg
Therefore, the acceleration is
(forward)
Learn more about forces, weight and Newton's second law:
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1 kilometre=1000 metre
1 hour = 3600 second
The initial velocity of car A is 35.0 km/hr i.e
= 9.72 m/s
The initial velocity of car B is 45 km/hr =12.5 m/s
The initial velocity of car C is 32 km/hr = 8.89 m/s
The initial velocity of car D is 110 km/hr=30.56 m/s
The acceleration of car A is given as
The time taken by car A = 15 min.
From equation of kinematics we know that-
v= u+at [Here v is the final velocity and a is the acceleration and t is the time]
Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s
=11.456111106 m/s
The acceleration of B is given as
The time taken by car B =20 min
The final velocity of B is -
v= u+at
= u-at [Here a is negative due to deceleration]
=12.5 m/s +[0.0011574074074×20×60]
=13.8888888.....
=13.9
The acceleration of C is given as
The time taken by car C =30 min
The final velocity of C is-
v = u+at
=8.89 m/s+[0.003086419753×30×60] m/s
=14.4455555555..m/s
=14.45 m/s
The car C is decelerating.The deceleration is given as-
The time taken by car D= 45 min.
The final velocity of the car D is-
v =u+at
=30.56 -[0.00462962962962×45×60]m/s
=18.06 m/s
Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.