What is the work done on the box by the weight of the box?

Asked by maham237 @ in Physics viewed by 282 People

A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ϕ=30∘ above the horizontal, as shown (Figure 6). The normal force acting on the box has a magnitude n=1.7N, the coefficient of kinetic friction between the box and the plane is μk=0.30, and the displacement d⃗ of the box is 1.8 m down the inclined plane.

Answered by maham237 @


Answer:

The work done is 0.882 Joules.

Explanation:

To calculate the work, we need to find all forces that are involved in the movement.

As you can analyse, the body is the force that make the box to move, and the friction force is opposite to it, we cannot forget about the friction. So, we have to calculate the resultant force to this context.

F_{R}=W_{x}-F_{k}

So, to find W_{x} we use: W_{x}=W.sen\phi; where W=2N;\phi = 30\°

W_{x}=2N.sin30\°=1\frac{1}{2}N=1N

The friction force would be:

F_{k}=\mu_{k}N=(0.30)(1.7N)=0.51

Then, the resultant force is:

F_{R}=W_{x}-F_{k}

F_{R}=1N-0.51N=0.49N

Now, we calculate the work: W=F_{R} d

W=(0.49N)(1.8m)=0.882J

Therefore, the work done is 0.882 Joules.


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