Answer:
Δv = 12 m/s, but we are not given the direction, so there are really an infinite number of potential solutions.
Maximum initial speed is 40.6 m/s
Minimum initial speed is 16.6 m/s
Explanation:
Assume this is a NET impulse so we can ignore friction.
An impulse results in a change of momentum
The impulse applied was
p = Ft = 1400(6.0) = 8400 N•s
p = mΔv
Δv = 8400 / 700 = 12 m/s
If the impulse was applied in the direction the car was already moving, the initial velocity was
vi = 28.6 - 12 = 16.6 m/s
if the impulse was applied in the direction opposite of the original velocity, the initial velocity was
vi = 28.6 + 12 = 40.6 m/s
Other angles of Net force would result in various initial velocities.
Answer:
Time, t = 0.87 seconds
Explanation:
Given that,
Initial velocity of the object, u = 4.3 m/s
The coefficient of kinetic friction between horizontal tabletop and the object is 0.5
We need to find the time taken by the object for the object to come to rest i.e. final velocity will be 0.
Using first equation of motion to find it as :
a is the acceleration, here, 
So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.
Answer:
A. The applied force should be the same size as the friction force
Explanation:
Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:
<u>A. The applied force should be the same size as the friction force</u>
