A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of 0.5 m/s2. What is the value of the friction force on the box?

200 N

50 N

43 N

33 N

F - F (friction) = m. a

50cos (30) - F (friction) = 20 x 0.5

F (frictions) = 50cos (30) - 10 = 33.3 N

Option - (D) : The value of the frictional force will be around 33 N.

Explanation:

So, we have a force, F of 50 N which is applied by a boy on a box of mass around 20 kg with an angle of 30 degree. We have the acceleration, a of 0.5 m/sec². As we know that each actions get some opposition and when we apply some force, F to an object there is a minimum amount of opposing force, F (friction) to it. And to find out the value of the frictional force, F (friction) we have the following equation to consider.

F (applied) - Friction = m. a, F*cos (angle) - friction (tot) = mass (m). acceleration (a), Now, putting all the required values inside the equation: 50*cos (30) - frictional force (F) = (20). (0.5), 50 * (0.866) - frictional force (F) = 10, 43.3-frictional force (F) = 10, frictional force (F) = 43.3-10, F=33.3 N,⇒Answer.