Answer:
a. Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.
b. -2.81 m/s²
c. 3.78 s
d. -351.25 N
Explanation:
a. After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.
For the block of ice to reach the top of the cliff, its maximum height, h should be greater than or equal to 5.7 m. That is, h ≥ 5.7 m.
The maximum height of a projection h, projected with an initial velocity v at an angle Ф is h = v²sin²Ф/2g where g = acceleration due to gravity = 9.8 m/s².
For the block of ice, v = 15 m/s and Ф = 45°. So,
h = v²sin²Ф/2g
= (15 m/s)²sin²45/(2 × 9.8 m/s²)
= 225 (m/s)²(1/√2)²/19.6 m/s²
= 225 (m/s)²(1/2)/19.6 m/s²
= 112.5 (m/s)²/19.6 m/s²
= 5.74 m
Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.
The graph is in the attachment.
b. What is the rate of acceleration while the blocks slow to a stop?
Using v² = u² + 2as where u = initial horizontal velocity of block = 15m/scos45° = 10.61 m/s, v = final velocity of block = 0 m/s since it stops, a = acceleration and s = distance block moves = 20 m
So, a = (v² - u²)/2s
substituting the variables into the equation, we have
a = ((0 m/s)² - (10.61 m/s)²)/2(20 m)
= - 112.57 (m/s)²)/40 m
= -2.81 m/s²
c. How long do the blocks take to slow to a stop?
Using v = u + at where u = initial horizontal velocity of block = 10.61 m/s v = final velocity of block = 0 m/s since it stops, a = acceleration = -2.81 m/s² and t = time it takes block of ice to stop
So, making t subject of the formula,
t = (v - u)/a
substituting the values of the variables, we have
t = ( 0 m/s - 10.61 m/s)/-2.81 m/s²
= -10.61 m/s/-2.81 m/s²
= 3.78 s
d. What is the amount of friction between the ice and the snowy ground?
The frictional force, f = net force on block of ice
f = ma where m = mass of bock = 125 kg and a = acceleration of block = -2.81 m/s²
f = ma
= 125 kg(2.81 m/s²)
= -351.25 N
