A 59.8 kg box initially at rest is pushed 2.88 m along a rough, horizontal floor with a constant applied horizontal force of 182.31 N. If the coefficient of friction between box and floor is 0.293, find the final speed of the box.
1 Answer | Add Yours
A 59.8 kg box that is initially at rest is pushed 2.88 m along a rough, horizontal floor with a constant applied horizontal force of 182.31 N. The coefficient of friction between the box and the floor is 0.293.
The work done by friction is equal to F = c*N, where c is the coefficient of friction and N is the normal force.
Here, the mass of the box is 59.8 kg. As the gravitational acceleration is 9.8 m/s^2, the normal force is 59.8*9.8 = 586.04 N. The frictional force acting between the box and the floor is 586.04*0.293 = 171.70 N. The constant force applied while the box was being pushed was 182.31 N. The frictional force acts in a direction opposite to that in which the box is pushed. This gives a net applied force of 182.31 - 171.7 = 10.61 N.
The force accelerates the box by 10.61/59.8 = 0.177 m/s^2. As the box is initially at rest and it is pushed for a distance equal to 2.88 m, the final speed V can be determined by using the formula V^2 - u^2 = 2*a*s where V is the final speed, u is the initial speed, s is the distance moved and a is the acceleration.
As u = 0, V^2 = 2*0.177*2.88 = 1.021
V = sqrt 1.021 = 1.01 m/s
The final speed of the box is 1.01 m/s
We’ve answered 318,979 questions. We can answer yours, too.Ask a question