At the beginning of a new school term, a
student moves a box of books by attaching a
rope to the box and pulling with a force of
F = 97.4 N at an angle of 62 degree, as shown in the
figure.
The acceleration of gravity is 9.8 m/s2 .
The box of books has a mass of 17 kg and
the coefficient of kinetic friction between the
bottom of the box and the floor is 0.32.
What is the acceleration of the box?
Answer in units of m/s2
student moves a box of books by attaching a
rope to the box and pulling with a force of
F = 97.4 N at an angle of 62 degree, as shown in the
figure.
The acceleration of gravity is 9.8 m/s2 .
The box of books has a mass of 17 kg and
the coefficient of kinetic friction between the
bottom of the box and the floor is 0.32.
What is the acceleration of the box?
Answer in units of m/s2
-
Hello AZs, let me give the formula in symbolic. Let F be the force applied at angle @.
Force used to pull F cos @
Normal force on the floor = Weight - F sin@ = Mg - F sin @
While moving the frictional force opposite to the applied force F cos@ will be u(Mg-Fsin@)
Hence net force = Fcos@ - u(Mg - Fsin@)
So acceleration a = [Fcos@ - u(Mg - Fsin@)] / M
Plug yourself F=97.4 N, @=62 deg, u = 0.32, M=17kg, g = 9.8 m/s^2
Solution: 1.17 m/s^2
Force used to pull F cos @
Normal force on the floor = Weight - F sin@ = Mg - F sin @
While moving the frictional force opposite to the applied force F cos@ will be u(Mg-Fsin@)
Hence net force = Fcos@ - u(Mg - Fsin@)
So acceleration a = [Fcos@ - u(Mg - Fsin@)] / M
Plug yourself F=97.4 N, @=62 deg, u = 0.32, M=17kg, g = 9.8 m/s^2
Solution: 1.17 m/s^2