A baseball of mass m1 = 0.45 kg is thrown at a concrete block m2 = 7.25 kg. The block has a coefficient of static friction of μs
1 answer:
Answer:
a.
b. 21.64 m/s
Explanation:
Let g = 9.81m/s2
a. The weight of the block is product of its mass and gravitational acceleration
which is also the normal force acting on the block from the floor so it stays balanced.
N = 71.225N
The static friction of the block is product of its normal force from the floor and the friction coefficient
For the block to move, the force generated by the impact must be at least equal to the static friction.
The impulse is product of this force and time duration of impact.
As impulse is generated by change in momentum of the ball, which is product of its mass and velocity v
b.
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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:
v = 0
Explanation:
This problem can be solved by taking into account:
- The equation for the calculation of the period in a spring-masss system
( 1 )
- The equation for the velocity of a simple harmonic motion
( 2 )
where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block
Hence
and by reeplacing it in ( 2 ):
In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.