Answer:
a) 90.9 kg
b) 22 m
Explanation:
(a) [tex]x-x_{0}=11 m[/tex]
t = 5.50s
[tex]v_{0x} =0[/tex]
[tex]x-x_{0}=[/tex][tex]v_{0x}[/tex]t+1/2*a_x*t^2
PLAN:
find the acceleration, then m by ∑Fx = m*a_x
Let +x be the direction of the force
∑Fx =80 N
a_x=2([tex]x-x_{0}[/tex])/t^2
=0.880m/s^2
m =∑Fx/a_x
= 90.9 kg
For (b). first calculate the speed at the end of the period (4.5 s) of applied force (accelerated kinematics) Then use this ending velocity as the initial (and constant) in the second part of motion
(b) a_x = 0 and v_x is constant.
After the first 5.0 s
v_x=[tex]v_{0x}[/tex]+a_x*t
=4.40 m/s
This is now the initial speed for the motion of the next 4.20 s (a=0)
[tex]x-x_{0}=[/tex] [tex]v_{0x}[/tex]t+1/2*a_x*t^2
= 22 m
note:
please verify the calculation
(a)107.2kg
(b)15.678
(a) Using Newton's second law of motion; i.e
∑F = m x a -------------(i)
Where;
∑F is the effective force
m = mass of object
a = acceleration
Since frictional force is negligible and the motion is strictly horizontal, the only force acting on the body is the applied horizontal force. i.e
∑F = 89.0N
Substitute this value into equation (i) as follows;
89.0 = m x a --------------------(ii)
Now let's calculate the acceleration of the block using one of the equations of motion as follows;
s = ut + (1/2)at² -----------------(iii)
Where;
s = horizontal distance = 12.5m
u = initial velocity = 0m/s (since the block starts from rest)
t = time taken for the part of motion = 5.50s
Substitute these values into equation (iii) as follows;
12.5 = (0)(5.50) + (1/2)(a)(5.50)²
12.5 = 0 + (1/2)(a)(30.25)
12.5 = 15.125a
a = 12.5/15.125 = 0.83m/s²
Substitute this value into equation (ii) as follows;
89.0 = m x (0.83)
m = 89.0/0.83
m = 107.2kg
Therefore, the mass of the block of ice is 107.2kg.
(b) Let's first get the velocity at the end of the first 4.5s as follows;
Using;
v = u + at; -------------------(iv)
Where;
v = velocity at 4.5s
u = initial velocity = 0 [since the block starts from rest]
a = acceleration = 0.83m/s²
Substitute these values into equation(iv) as follows;
v = 0 + (0.83) x 4.5
v = 3.735m/s [This is the velocity for the first 4.5seconds]
Now, lets get the distance moved by the block in the next 4.20s
Now the velocity is constant (i.e not changing). Therefore, acceleration is zero.
i.e
a = 0
Using equation (iii)
Where;
u = 3.735m/s
t = 4.2s
Substitute these values;
s = ut + (1/2)at²
s = 3.735(4.2) + 0
s = 15.678m
Therefore the distance moved by the block in the next 4.20s is 15.678m
