A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surfac
ID: 2127367 • Letter: A
Question
A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal
surface.
The mass is initially at rest and covers a distance of 5 m in 0.92 s
under the action of the force.
Assuming there are no energy losses due to
air resistance and therefore that the acceleration is constant:
Calculate the total energy expended in the acceleration.
Plot a graph of the Kinetic energy of the mass against time
Plot a graph of the kinetic energy of the mass against distance
Calculate the coefficient of friction between the mass and the surface
Explanation / Answer
If there were no friction, teh force F would do work W given by:
W = Fd where d = distance the force acts over so using your numbers
W = 80N*5m = 400 J (a)
Now it takes the box 0.92s to move the 5m and you are told acceleration is constant so you can use
d = 1/2 at^2 to find a ---> a = 2d/t^2 = 2*5 m/(0.92s)^2 = 11.82 m/s^2
The net force, the difference between the applied force F and friction, is then:
Fnet = ma = 70.92 =N
Then,
Fnet = F - umg where u = coefficient of friction and g = 9.8 m/s^2
SOlving for u ---> u = (F - Fnet)/mg = 0.154