Question

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

Answer 1

F - F(friction) = m.a
50cos(30) - F(friction) = 20 x 0.5
F(frictions) = 50cos(30) - 10 = 33.3 N

Answer 2

Answer:

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,

1. F×cos(angle)-friction(tot)=mass(m).acceleration(a),
2. Now, putting all the required values inside the equation:
3. 50×cos(30)-frictional force(F)=(20).(0.5),
4. 50×(0.866)-frictional force(F)=10,
5. 43.3-frictional force(F)=10,
6. frictional force(F)=43.3-10,
7. F=33.3 N,⇒Answer.