Question

A box, mass of 10kg, is being pulled to the right with a force of 100N at an angle of 45 degrees above the horizontal. The coefficient of kinetic friction between the box and the floor is 0.2 , what is the acceleration of the box?

Answer 1

Answer on Question #64204, Physics / Mechanics | Relativity

A box, mass of $10\mathrm{kg}$ , is being pulled to the right with a force of 100N at an angle of 45 degrees above the horizontal. The coefficient of kinetic friction between the box and the floor is 0.2, what is the acceleration of the box?

Solution:

$$m a = F _ {a x} - F _ {f}$$ $$F _ {a x} = F _ {a} \cos 4 5 {}^ {\circ} = (1 0 0 \mathrm {N}) \cos 4 5 {}^ {\circ} = 7 0. 7 1 \mathrm {N}$$

The friction force is

$$F _ {f} = \mu F _ {n}$$

The normal force is

$$F _ {n} = m g - F _ {a y}$$

where

$$F _ {a y} = F _ {a} \sin 4 5 {}^ {\circ} = (1 0 0 \mathrm {N}) \sin 4 5 {}^ {\circ} = 7 0. 7 1 \mathrm {N}$$

So,

$$F _ {f} = \mu (m g - F _ {a y}) = 0. 2 ((1 0 k g) (9. 8 0 m / s ^ {2}) - 7 0. 7 1 N) = 5. 4 6 N$$

Thus, the acceleration is

$$a = \frac {F _ {a x} - F _ {f}}{m} = \frac {7 0 . 7 1 - 5 . 4 6}{1 0} = 6. 5 2 5 \mathrm {m / s ^ {2}}$$

Answer: $6.525 \, \text{m/s}^2$

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