Question

8. Two forces of 10 N and 30 N are applied to a 10 kg box. Find (1) the box’s acceleration when both forces point due east and (2) the box’s acceleration when 10 N force points due east and 30 N force points due west.

Answer 1

(1) acceleration, a = 4 m/
`s^(2)` (2) acceleration of 10 N,
`a_(1)` = 1 m/
`s^(2)` and acceleration of 30 N,
`a_(2)` = 3 m/
`s^(2)`

Step-by-step explanation:

• Here, the acceleration of the object could be found using the equation derived in the second law of motion. The equation is given as, F = ma where m is the acceleration of the object, m is the mass of the object and F is the applied on the object.
• Let
  `a_(1)` be the acceleration for force 10 N, to find acceleration rearrange the equation to a =
  `(F)/(m)`. When we substitute 10 N force and 10 kg mass of the box in the equation. We will get
  `a_(1)` = 1 m/
  `s^(2)`

• Let
  `a_(2)` be the acceleration for force 30 N, to find acceleration rearrange the equation to F =
  `(F)/(m)`. When we substitute 30 N force and 10 kg mass of the box in the equation. We will get
  `a_(2)` = 3 m/
  `s^(2)`
• To find the combined, just add the force and substitute in the above equation. Hence, a = 4 m/
  `s^(2)`