If the surface is rough then the average force needed to move the wooden block will be more as the friction between the surfaces will be more.
Hence, the least force will be required on the smoothest surface and the greatest force will be required for the roughest surface.
a) Hence, the order will be: B > D > A > C
b) When the surface is glued then the force of friction will increase. Hence, it will require more force to move the wooden block on the surface. Hence, the more required will be more than 105 N.
<span>When the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.
Given:
a = 86 m/s^2
t = 1.7 s
Solution:
d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.7) + 0.5 (86) (1.7)^2
d = 124.27 m
vf = vi + at
vf = 0 + 86 (1.7)
vf = 146.2 m/s (velocity when the fuel is consumed completely)
Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 146.2 + (-9.8) (t)
t = 14.92 s
Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2)
d = 1090.53 m
Then, we determine the maximum altitude:
d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m</span>
Answer: A
Hope this help you!!
Answer:
x = 25 / μ [ ft]
Explanation:
To solve this exercise we can use Newton's second law.
Let's set a reference system where the x axis is parallel to the road
Y axis
N_B + N_A - W_van - W_load = 0
N_B + N_A = W_van + W_load
X axis
fr = ma
a = fr / m
the total mass is
m = (W_van + W_load) / g
the friction force has the expression
fr = μ N_{total}
fr = μy (W_van + W_load)
we substitute
a = μ (W_van + W_load)
a = μ g
taking the acceleration let's use the kinematic relations where the final velocity is zero
v² = v₀² - 2 a x
0 = v₀² -2a x
x =
x =
x =
x = 25 / μ [ ft]