<span>In this particular case, where car is moving through curvature, so it is moving in circular motion, force acting on car is centripetal force which holds car not to fly out. Centripetal force is always pointed in the middle of circle. Here frictional force has role of centripetal force. If frictional force is to weak, car would fly out of curvutare.</span>
Answer:
3.71 m/s in the negative direction
Explanation:
From collisions in momentum, we can establish the formula required here which is;
m1•u1 + m2•v2 = m1•v1 + m2•v2
Now, we are given;
m1 = 1.5 kg
m2 = 14 kg
u1 = 11 m/s
v1 = -1 m/s (negative due to the negative direction it is approaching)
u2 = -5 m/s (negative due to the negative direction it is moving)
Thus;
(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)
This gives;
16.5 - 70 = -1.5 + 14v2
Rearranging, we have;
16.5 + 1.5 - 70 = 14v2
-52 = 14v2
v2 = - 52/14
v2 = 3.71 m/s in the negative direction
Answer:
P= 454.11 N
Explanation:
Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).
The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:
In order for the small cube to not slide down, the friction force must equal the weight of the small cube:
The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N
