When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
Answer:
Initial velocity, U = 4.5m/s
Explanation:
Given the following data;
Final velocity, v = 12m/s
Time, t = 5 seconds
Acceleration, a = 1.5m/s²
To find the initial velocity, we would use the first equation of motion.
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the equation, we have;
12 = U + 1.5*5
12 = U + 7.5
U = 12 - 7.5
Initial velocity, U = 4.5m/s
B. F<em>spring = k(triangle)</em> x
Answer:
The force of the impact would be smaller
Explanation: Examples:
If the force is big then the time would be small (2500N of Force = 10 seconds)
If the force is small then the time would be big (250N of Force = 50 seconds)
Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum
The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.
-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.
-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.
-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic.
