A 10.0 kg box is dragged along a horizontal floor by a cord that is pulled upward at an angle of 30 degrees above the horizontal as shown below. The tension in the cord is 100 N and the coefficient of kinetic friction between the box and floor is 0.500. What is the acceleration of the box (in m/s^2)? Here is a diagram of the problem:

Question

A 10.0 kg box is dragged along  a horizontal floor by a cord that is pulled upward at an angle of 30 degrees above the horizontal as shown below. The tension in the cord is 100 N and the coefficient of kinetic friction between the box and floor is 0.500. What is the acceleration of the box (in m/s^2)? Here is a diagram of the problem:

anonymous · Answer

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anonymous · Answer

First, we need to work out the horizontal component of the tension force:

$$\cos30 = Th / 100$$
$$Th = 100\cos30$$
$$Th \approx 86.6 N$$

We know that

$$μ = F / N$$
So the backwards force from friction is

$$0.5 = F / (10 * 9.8)$$
$$F = 49 N$$

Now we need the total horizontal force acting on the box, which is

$$86.6 - 49 = 37.6 N$$

Now that we have the net horizontal force and also the mass of the box, we have everything we need to find the acceleration of the box:

$$F = ma$$
$$a = F / m$$
$$\approx 37.6 / 10$$
$$\approx 3.76 m/s^2$$

Normally with physics problems it's better to keep everything in equation form until the very end to avoid rounding errors and such. I didn't do that in this case because I think using the numbers makes it easier to see what's going on. This is why there are some 'approximately equal' signs.